CBSE Class 12 Mathematics Probability Assignment Set 04

Selected NCERT Questions

Question. A black and red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9 given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8 given that the red die resulted in a number less than 4.
Answer: When a black and a red die are rolled then \( n(S) = 36 \).
(a) Let A be the event getting sum greater than 9 and B be the event getting a 5 on the black die.
\( \therefore \) \( A = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\} \)
\( B = \{(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\} \)
\( A \cap B = \{(5, 5), (5, 6)\} \)
\( \therefore \) \( P(A) = \frac{6}{36} = \frac{1}{6} \), \( P(B) = \frac{6}{36} = \frac{1}{6} \), and \( P(A \cap B) = \frac{2}{36} = \frac{1}{18} \)
\( \therefore \) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{18}}{\frac{1}{6}} = \frac{1}{18} \times \frac{6}{1} = \frac{1}{3} \)
(b) Let A be the event getting the sum 8 and B be the event getting a number less than 4 on red die.
\( A = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \)
\( B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)\} \)
\( A \cap B = \{(2, 6), (3, 5)\} \)
\( \therefore \) \( P(A) = \frac{5}{36} \), \( P(B) = \frac{18}{36} = \frac{1}{2} \), \( P(A \cap B) = \frac{2}{36} = \frac{1}{18} \)
\( \therefore \) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{18}}{\frac{1}{2}} = \frac{1}{18} \times \frac{2}{1} = \frac{1}{9} \)

Question. An instructor has a question bank consisting of 300 easy true/false questions, 200 difficult, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Answer: Here, total questions = \( 300 + 200 + 500 + 400 = 1400 \)
Let A be the event that selected question is an easy question.
\( \therefore \) \( P(A) = \frac{300 + 500}{1400} = \frac{800}{1400} = \frac{4}{7} \)
Let B be the event that selected question is a multiple choice question.
\( P(B) = \frac{500 + 400}{1400} = \frac{900}{1400} = \frac{9}{14} \)
Now \( A \cap B \) is the event so that the selected question is a easy multiple choice question.
\( \therefore \) \( P(A \cap B) = \frac{500}{1400} = \frac{5}{14} \)
\( \therefore \) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{5}{14}}{\frac{9}{14}} = \frac{5}{9} \)

Question. Consider the experiment of throwing a die. If a multiple of 3 comes up, a die is again thrown and if any other number comes, a coin is tossed. Find the conditional probability of the event, 'the coin shows a tail' given that 'at least one die shows a 3'.
Answer: Here, \( S = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T)\} \)
Let A be the event of getting a tail on coin.
\( A = \{(1, T), (2, T), (4, T), (5, T)\} \)
Let B be the event of getting 3 on at least one die.
\( B = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 3)\} \)
\( \therefore \) \( A \cap B = \phi \)
\( \therefore \) \( P(A) = \frac{4}{20} = \frac{1}{5} \), \( P(B) = \frac{7}{20} \) and \( P(A \cap B) = \frac{0}{20} = 0 \)
\( \therefore \) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{\frac{7}{20}} = 0 \)

Question. A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.
Answer: We have, \( P(A) = \frac{1}{2} \) and \( P(B) = \frac{1}{6} \)
Also, \( P(A \cap B) = P(\text{head appears on coin and 3 on the die}) = \frac{1}{12} \)
Clearly, \( P(A \cap B) = P(A) \times P(B) \)
Thus, A and B are independent events.

Question. Events A and B are such that \( P(A) = \frac{1}{2} \), \( P(B) = \frac{7}{12} \) and \( P(\text{not A or not B}) = \frac{1}{4} \). State whether A and B are independent.
Answer: Here \( P(A) = \frac{1}{2} \), \( P(B) = \frac{7}{12} \) and \( P(\overline{A} \cup \overline{B}) = \frac{1}{4} \)
Now \( P(\overline{A} \cup \overline{B}) = P(\overline{A \cap B}) = 1 - P(A \cap B) \)
\( \frac{1}{4} = 1 - P(A \cap B) \)
\( \implies \) \( P(A \cap B) = 1 - \frac{1}{4} = \frac{3}{4} \)
Now \( P(A) \times P(B) = \frac{1}{2} \times \frac{7}{12} = \frac{7}{24} \)
\( \therefore \) \( P(A \cap B) \neq P(A) \times P(B) \)
Thus, A and B are not independent.

Question. Probabilities of solving specific problem independently by A and B are \( \frac{1}{2} \) and \( \frac{1}{3} \) respectively. If both try to solve the problem independently. Find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.
Answer: Here, \( P(A) = \frac{1}{2} \) and \( P(B) = \frac{1}{3} \)
Now \( P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \), \( P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \)
(i) \( P(\text{the problem is solved}) = 1 - P(\overline{A} \cap \overline{B}) \)
\( = 1 - P(\overline{A}) P(\overline{B}) \)
\( = 1 - \left( \frac{1}{2} \times \frac{2}{3} \right) = 1 - \frac{1}{3} = \frac{2}{3} \)
(ii) \( P(\text{exactly one of them solves}) = P(A) P(\overline{B}) + P(\overline{A}) P(B) \)
\( = \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3} = \frac{2}{6} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \).

Question. In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspaper. A student is selected at random.
(a) Find the probability that the student reads neither Hindi nor English newspaper.
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Answer: Let A be the event that a student reads Hindi newspaper and B be the event that a student reads English newspaper.
\( P(A) = \frac{60}{100} = 0.6 \), \( P(B) = \frac{40}{100} = 0.4 \) and \( P(A \cap B) = \frac{20}{100} = 0.2 \)
(a) Now \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( = 0.6 + 0.4 - 0.2 = 0.8 \)
Probability that she reads neither Hindi nor English newspaper
\( = 1 - P(A \cup B) = 1 - 0.8 = 0.2 = \frac{1}{5} \)
(b) \( P(B/A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.6} = \frac{1}{3} \)
(c) \( P(A/B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.4} = \frac{1}{2} \)

Question. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Answer: Let \( E_1 \) and \( E_2 \) be the events that red ball is drawn in first draw and black ball is drawn in first draw respectively. Let A be the event that ball drawn in second draw is red. There are 5 red and 5 black balls in the urn.
\( \therefore \) \( P(E_1) = \frac{5}{10} = \frac{1}{2} \) and \( P(E_2) = \frac{5}{10} = \frac{1}{2} \)
When 2 additional balls of red colour are put in the urn there are 7 red and 5 black balls in the urn.
\( \therefore \) \( P\left(\frac{A}{E_1}\right) = \frac{7}{12} \)
When 2 additional balls of black colour are put in the urn there are 5 red and 7 black balls in the urn.
\( \therefore \) \( P\left(\frac{A}{E_2}\right) = \frac{5}{12} \)
By theorem of total probability
\( P(A) = P(E_1) P\left(\frac{A}{E_1}\right) + P(E_2) P\left(\frac{A}{E_2}\right) \)
\( \implies \) \( P(A) = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} = \frac{7}{24} + \frac{5}{24} = \frac{12}{24} = \frac{1}{2} \)

Question. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade. What is the probability that the student is a hostlier?
Answer: Let the event be defined as
\( E_1 \) = selection of hostelier
\( E_2 \) = selection of day scholar
A = selection of student getting A grade
\( P(E_1) = \frac{60}{100} = \frac{3}{5} \), \( P(E_2) = \frac{40}{100} = \frac{2}{5} \)
\( P\left(\frac{A}{E_1}\right) = \frac{30}{100} = \frac{3}{10} \), \( P\left(\frac{A}{E_2}\right) = \frac{20}{100} = \frac{1}{5} \)
\( P\left(\frac{E_1}{A}\right) \) = required
\( P\left(\frac{E_1}{A}\right) = \frac{P(E_1) \cdot P\left(\frac{A}{E_1}\right)}{P(E_1) \cdot P\left(\frac{A}{E_1}\right) + P(E_2) \cdot P\left(\frac{A}{E_2}\right)} = \frac{\frac{3}{5} \cdot \frac{3}{10}}{\frac{3}{5} \cdot \frac{3}{10} + \frac{2}{5} \cdot \frac{1}{5}} = \frac{\frac{9}{50}}{\frac{9}{50} + \frac{2}{25}} = \frac{\frac{9}{50}}{\frac{9+4}{50}} = \frac{9}{13} \)

Question. A Laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested then with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease then what is the probability that a person has the disease given that his test result is positive.
Answer: Let \( E_1 \) and \( E_2 \) denote the events that a person has disease and does not have disease respectively. Let A be the event that the test result is positive.
Now, the probability that a person has the disease is
\( P(E_1) = 0.1\% = \frac{0.1}{100} = 0.001 \)
Probability that a person does not have the disease
\( \therefore \) \( P(E_2) = 1 - 0.001 = 0.999 \)
Probability that a person has disease and test result is positive.
\( \therefore \) \( P(A/E_1) = 99\% = \frac{99}{100} = 0.99 \)
Probability that a person does not have disease and test result is positive.
\( \therefore \) \( P(A/E_2) = 0.5\% = \frac{0.5}{100} = 0.005 \)
By Bayes' theorem,
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.00099 + 0.004995} \)
\( = \frac{0.00099}{0.005985} = \frac{990}{5985} = \frac{22}{133} \)

Question. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer: Let \( E_1, E_2, E_3 \) and \( E_4 \) be the events that the missing card is a heart, spade, club and diamond respectively. Let A be the event of drawing two diamond cards from 51 cards.
There are four events and each event is equally likely to be selected.
\( \therefore \) \( P(E_1) = P(E_2) = P(E_3) = P(E_4) = \frac{13}{52} = \frac{1}{4} \)
It is given that
\( P(A/E_1) = \frac{{}^{13}C_2}{{}^{51}C_2} \), \( P(A/E_2) = \frac{{}^{13}C_2}{{}^{51}C_2} \), \( P(A/E_3) = \frac{{}^{13}C_2}{{}^{51}C_2} \) and \( P(A/E_4) = \frac{{}^{12}C_2}{{}^{51}C_2} \)
By Bayes' theorem,
\( P(E_4/A) = \frac{P(E_4)P(A/E_4)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3) + P(E_4)P(A/E_4)} \)
\( = \frac{\frac{1}{4} \times \frac{{}^{12}C_2}{{}^{51}C_2}}{\frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{13}C_2}{{}^{51}C_2} + \frac{1}{4} \times \frac{{}^{12}C_2}{{}^{51}C_2}} = \frac{{}^{12}C_2}{{}^{13}C_2 + {}^{13}C_2 + {}^{13}C_2 + {}^{12}C_2} = \frac{{}^{12}C_2}{3 \cdot {}^{13}C_2 + {}^{12}C_2} \)
\( = \frac{\frac{12!}{2!10!}}{3 \times \frac{13!}{2!11!} + \frac{12!}{2!10!}} = \frac{66}{3 \times 78 + 66} = \frac{66}{300} = \frac{11}{50} \)

Question. A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
Answer: Let X denote the random variable which denotes the number of tails when a biased coin is tossed twice.
So, X may have values 0, 1 or 2.
Since the coin is biased in which head is 3 times as likely to occur as a tail.
\( \therefore \) \( P(H) = \frac{3}{4} \) and \( P(T) = \frac{1}{4} \)
Now, \( P(X = 0) \) \( \implies \) Probability of getting two heads
\( P(X = 0) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \)
\( P(X = 1) \) \( \implies \) Probability of getting one tail and one head
\( P(X = 1) = \frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4} = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8} \)
\( P(X = 2) \) \( \implies \) Probability of getting two tails
\( P(X = 2) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \)
Thus, required probability distribution is

Question. The random variable X has a probability distribution P(X) of the following form, where k is some number:
\( P(X) = \begin{cases} k & \text{if } x = 0 \\ 2k & \text{if } x = 1 \\ 3k & \text{if } x = 2 \\ 0 & \text{otherwise} \end{cases} \)
(a) Determine the value of k.
(b) Find \( P(X < 2), P(X \leq 2), P(X \geq 2) \).
Answer: (a) \( k + 2k + 3k = 1 \) [\( \because p_1 + p_2 + p_3 + ... + p_n = 1 \)]
\( \implies \) \( 6k = 1 \) \( \implies \) \( k = \frac{1}{6} \)
(b) \( P(X < 2) = k + 2k = 3k = 3 \times \frac{1}{6} = \frac{1}{2} \)
\( P(X \leq 2) = k + 2k + 3k = 6k = 6 \times \frac{1}{6} = 1 \)
\( P(X \geq 2) = 3k = 3 \times \frac{1}{6} = \frac{1}{2} \)

Question. How many times should a man toss a fair coin so that the probability of having at least one head is more than 90%?
Answer: Let the coin be tossed n times.
Probability of getting a head = \( \frac{1}{2} \)
Probability of getting no head = \( \frac{1}{2} \)
Probability of getting atleast one head in n tosses = \( 1 - \left( \frac{1}{2} \right)^n \)
But probability of getting at least one head is more than 90% = \( \frac{90}{100} = \frac{9}{10} \)
\( \therefore \) \( 1 - \left( \frac{1}{2} \right)^n > \frac{9}{10} \)
\( \implies \) \( \left( \frac{1}{2} \right)^n < 1 - \frac{9}{10} \)
\( \implies \) \( \left( \frac{1}{2} \right)^n < \frac{1}{10} \)
\( \implies \) \( n \geq 4 \).

Question. Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.
Answer: Let \( E_1, E_2, A \) be events defined as
\( E_1 \) = treatment of heart attack with Yoga and meditation
\( E_2 \) = treatment of heart attack with certain drugs
A = person getting heart attack
\( P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2} \)
Now \( P\left(\frac{A}{E_1}\right) = 40\% - \left( 40 \times \frac{30}{100} \right) \% = 40\% - 12\% = 28\% = \frac{28}{100} \)
\( P\left(\frac{A}{E_2}\right) = 40\% - \left( 40 \times \frac{25}{100} \right) \% = 40\% - 10\% = 30\% = \frac{30}{100} \)
We have to find \( P\left(\frac{E_1}{A}\right) \)
\( \therefore \) \( P\left(\frac{E_1}{A}\right) = \frac{P(E_1).P\left(\frac{A}{E_1}\right)}{P(E_1).P\left(\frac{A}{E_1}\right) + P(E_2).P\left(\frac{A}{E_2}\right)} = \frac{\frac{1}{2} \times \frac{28}{100}}{\frac{1}{2} \times \frac{28}{100} + \frac{1}{2} \times \frac{30}{100}} = \frac{28}{100} \times \frac{100}{58} = \frac{14}{29} \)

Question. Bag I contains 3 red and 4 black balls and bag II contains 4 red and 5 black balls. One ball is transferred from bag I to bag II and then a ball is drawn from bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
Answer: Let \( E_1 \) = Event that a red ball is drawn from bag I
\( E_2 \) = Event that a black ball is drawn from bag I
Therefore, \( P(E_1) = \frac{3}{7}, P(E_2) = \frac{4}{7} \)
After transferring a red ball from bag I to bag II, the bag II will have 5 red and 5 black balls.
Let A be the event of drawing red ball
\( \therefore \) \( P(A/E_1) = \frac{5}{10} = \frac{1}{2} \)
Further, when a black ball is transferred from bag I to bag II, it will contain 4 red and 6 black balls.
\( P(A/E_2) = \frac{4}{10} = \frac{2}{5} \)
By Bayes' theorem, we have
\( P(E_2/A) = \frac{P(E_2).P(A/E_2)}{P(E_1).P(A/E_1) + P(E_2).P(A/E_2)} \)
\( = \frac{\frac{4}{7} \times \frac{2}{5}}{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}} = \frac{\frac{8}{35}}{\frac{3}{14} + \frac{8}{35}} = \frac{8}{35} \times \frac{70}{31} = \frac{16}{31} \)

Topic 12

Probability

Assignments

(i) Conditional Probability

Level I

Question 1. If P(A) = 0.3, P(B) = 0.2, find P(B/A) if A and B are mutually exclusive events.
Answer:
Since the events \( A \) and \( B \) are mutually exclusive, they cannot happen at the same time. This means their intersection is empty:
\( P(A \cap B) = 0 \)
By the definition of conditional probability, we have:
\( P(B/A) = \frac{P(A \cap B)}{P(A)} \)
Substituting the values gives:
\( P(B/A) = \frac{0}{0.3} = 0 \)
Therefore, the conditional probability \( P(B/A) \) is \( 0 \).
In simple words: Since the two events cannot occur together, if we know that event \( A \) has already happened, there is absolutely zero chance of event \( B \) happening.

Exam Tip: Always remember that the probability of the intersection of two mutually exclusive events is always zero. This is a fundamental concept frequently tested in board exams.

Question 2. Find the probability of drawing two white balls in succession from a bag containing 3 red and 5 white balls respectively, the ball first drawn is not replaced.
Answer:
The total number of balls in the bag initially is \( 3 + 5 = 8 \), and the number of white balls is \( 5 \).
The probability of selecting a white ball on the first draw is:
\( P(W_1) = \frac{5}{8} \)
Since the ball is not replaced, the total number of remaining balls becomes \( 7 \), and the number of remaining white balls becomes \( 4 \).
The probability of selecting a white ball on the second draw, given that the first ball was white, is:
\( P(W_2/W_1) = \frac{4}{7} \)
By the multiplication rule of probability, the probability of drawing two white balls in succession is:
\( P(W_1 \cap W_2) = P(W_1) \cdot P(W_2/W_1) \)
\( P(W_1 \cap W_2) = \frac{5}{8} \cdot \frac{4}{7} = \frac{5}{14} \)
Thus, the required probability is \( \frac{5}{14} \).
In simple words: First find the probability of drawing a white ball from the full bag (5 out of 8). Next, find the probability of getting another white ball from the remaining ones (4 out of 7). Multiply these fractions together to get the final answer.

Exam Tip: In non-replacement questions, always remember to decrease both the numerator and the denominator by 1 for the second trial.

Level II

Question 1. A dice is thrown twice and sum of numbers appearing is observed to be 6. what is the conditional probability that the number 4 has appeared at least once.
Answer:
Let \( F \) be the event that the sum of the numbers appearing on the two dice is \( 6 \). The outcomes in \( F \) are:
\( F = \{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\} \)
The total number of outcomes in \( F \) is \( n(F) = 5 \).
Let \( E \) be the event that the number \( 4 \) appears at least once on either throw:
\( E = \{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)\} \)
The intersection \( E \cap F \) represents outcomes where the sum is 6 and 4 appears at least once:
\( E \cap F = \{(2, 4), (4, 2)\} \)
The number of outcomes in \( E \cap F \) is \( n(E \cap F) = 2 \).
The required conditional probability \( P(E/F) \) is:
\( P(E/F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{5} \)
Thus, the conditional probability is \( \frac{2}{5} \).
In simple words: First write down all the possible ways to get a sum of 6, which gives 5 combinations. Out of these 5 ways, look for the ones that contain the number 4, which is only 2 combinations. The probability is then 2 out of 5.

Exam Tip: Listing the elements of the sample spaces for conditional events prevents errors and helps the evaluator follow your solution steps easily.

Level III

Question 1. If P(A) = 3/8 , P(B) = 1/2 and P(A \(\cap\) B) = 1/4 , find P(\(\bar{A}\)/\(\bar{B}\)) and P(\(\bar{B}\)/\(\bar{A}\))
Answer:
We first find the probability of the union of events \( A \) and \( B \):
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
\( P(A \cup B) = \frac{3}{8} + \frac{1}{2} - \frac{1}{4} = \frac{3 + 4 - 2}{8} = \frac{5}{8} \)
Using De Morgan's Law, the probability that neither \( A \) nor \( B \) occurs is:
\( P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - \frac{5}{8} = \frac{3}{8} \)
We also need the individual complement probabilities:
\( P(\bar{A}) = 1 - P(A) = 1 - \frac{3}{8} = \frac{5}{8} \)
\( P(\bar{B}) = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2} \)
Now, the required conditional probabilities are:
\( P(\bar{A}/\bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} = \frac{3/8}{1/2} = \frac{3}{4} \)
\( P(\bar{B}/\bar{A}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{A})} = \frac{3/8}{5/8} = \frac{3}{5} \)
Therefore, \( P(\bar{A}/\bar{B}) = \frac{3}{4} \) and \( P(\bar{B}/\bar{A}) = \frac{3}{5} \).
In simple words: First find the probability that at least one event happens, which is 5/8. Its opposite (that neither happens) is 3/8. Use this to compute the conditional complement probabilities by dividing by the respective individual complements.

Exam Tip: Clearly show the use of De Morgan's Law \( P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) \), as this formula is highly valued in the marking scheme.

(ii) Multiplication Theorem on Probability

Level II

Question 1. A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find what is the probability that none is red.
Answer:
The total number of balls in the bag is \( 5 + 7 + 3 = 15 \).
The number of red balls is \( 7 \), and the number of non-red (white or black) balls is \( 5 + 3 = 8 \).
We want to draw three balls, none of which are red:
- Probability that the first ball is non-red: \( P(N_1) = \frac{8}{15} \)
- Probability that the second ball is non-red, given the first was non-red: \( P(N_2/N_1) = \frac{7}{14} = \frac{1}{2} \)
- Probability that the third ball is non-red, given the first two were non-red: \( P(N_3/N_1 \cap N_2) = \frac{6}{13} \)
Using the multiplication theorem of probability, the required probability is:
\( P(\text{none is red}) = \frac{8}{15} \cdot \frac{1}{2} \cdot \frac{6}{13} = \frac{24}{195} = \frac{8}{65} \)
Thus, the probability that none of the balls drawn is red is \( \frac{8}{65} \).
In simple words: Since we want no red balls, we only draw from the 8 white and black balls. Multiply the decreasing probabilities for each step: 8/15 for the first, 7/14 for the second, and 6/13 for the third.

Exam Tip: Be sure to simplify fractions during the intermediate steps (like reducing \( \frac{7}{14} \) to \( \frac{1}{2} \)) to make the final calculation easier.

Question 2. The probability of A hitting a target is 3/7 and that of B hitting is 1/3. They both fire at the target. Find the probability that (i) at least one of them will hit the target, (ii) Only one of them will hit the target.
Answer:
The probabilities of hitting the target are given as:
\( P(A) = \frac{3}{7} \implies P(\bar{A}) = 1 - \frac{3}{7} = \frac{4}{7} \)
\( P(B) = \frac{1}{3} \implies P(\bar{B}) = 1 - \frac{1}{3} = \frac{2}{3} \)
Since the two events are independent, we calculate the required probabilities:

(i) Probability that at least one hits:
This is best solved by taking the complement of the probability that neither hits:
\( P(\text{at least one}) = 1 - P(\bar{A}) \cdot P(\bar{B}) \)
\( P(\text{at least one}) = 1 - \left( \frac{4}{7} \cdot \frac{2}{3} \right) = 1 - \frac{8}{21} = \frac{13}{21} \)

(ii) Probability that only one hits:
This occurs if A hits and B misses, or A misses and B hits:
\( P(\text{only one}) = P(A) \cdot P(\bar{B}) + P(\bar{A}) \cdot P(B) \)
\( P(\text{only one}) = \left( \frac{3}{7} \cdot \frac{2}{3} \right) + \left( \frac{4}{7} \cdot \frac{1}{3} \right) = \frac{6}{21} + \frac{4}{21} = \frac{10}{21} \)
Therefore, the probability that at least one hits is \( \frac{13}{21} \), and that only one hits is \( \frac{10}{21} \).
In simple words: (i) Find the chance that both miss (8/21) and subtract it from 1 to get the chance that at least one hits. (ii) Add the separate probabilities of A hitting while B misses, and B hitting while A misses.

Exam Tip: Using the complement method \( 1 - P(\text{none}) \) is much faster and less prone to errors than summing up all possible successful cases.

Level III

Question 1. A class consists of 80 students; 25 of them are girls and 55 are boys, 10 of them are rich and the remaining poor; 20 of them are fair complexioned. what is the probability of selecting a fair complexioned rich girl.
Answer:
Let us assume that the three characteristics (gender, wealth, and complexion) are independent of each other.
Let \( G \) be the event of choosing a girl, \( R \) be the event of choosing a rich student, and \( F \) be the event of choosing a fair-complexioned student.
The individual probabilities are:
\( P(G) = \frac{25}{80} = \frac{5}{16} \)
\( P(R) = \frac{10}{80} = \frac{1}{8} \)
\( P(F) = \frac{20}{80} = \frac{1}{4} \)
Since the events are independent, the probability of selecting a fair-complexioned rich girl is the product of their individual probabilities:
\( P(G \cap R \cap F) = P(G) \cdot P(R) \cdot P(F) \)
\( P(G \cap R \cap F) = \frac{5}{16} \cdot \frac{1}{8} \cdot \frac{1}{4} = \frac{5}{512} \)
Thus, the required probability is \( \frac{5}{512} \).
In simple words: Since we assume gender, wealth, and complexion are independent, find the fraction of girls (5/16), rich students (1/8), and fair students (1/4) in the class. Multiply these three fractions to get the final chance.

Exam Tip: When joint information is not given, we assume independence among different categories and simply multiply their individual probabilities.

Question 2. Two integers are selected from integers 1 through 11. If the sum is even, find the probability that both the numbers are odd.
Answer:
The integers are \( \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} \). Among these:
- Odd numbers: \( \{1, 3, 5, 7, 9, 11\} \) (6 numbers)
- Even numbers: \( \{2, 4, 6, 8, 10\} \) (5 numbers)
Let \( S \) be the event that the sum of the two selected numbers is even. The sum of two numbers is even if both are odd or both are even.
The total number of ways to choose 2 numbers out of 11 is:
\( n(\text{Total}) = C(11,2) = 55 \)
The number of ways to choose 2 odd numbers is:
\( n(\text{Both Odd}) = C(6,2) = 15 \)
The number of ways to choose 2 even numbers is:
\( n(\text{Both Even}) = C(5,2) = 10 \)
Thus, the total favorable outcomes for the sum to be even are:
\( n(S) = 15 + 10 = 25 \)
Let \( E \) be the event that both selected numbers are odd. The intersection \( E \cap S \) represents both numbers being odd, so:
\( n(E \cap S) = 15 \)
The required conditional probability \( P(E/S) \) is:
\( P(E/S) = \frac{n(E \cap S)}{n(S)} = \frac{15}{25} = \frac{3}{5} \)
Therefore, the probability is \( \frac{3}{5} \).
In simple words: A sum is even only when we pick two odd numbers (15 ways) or two even numbers (10 ways), giving 25 total possibilities. Out of these, we want both to be odd (15 ways), so the probability is 15/25, which simplifies to 3/5.

Exam Tip: Be sure to restrict the sample space to the given condition (sum is even) first, which makes solving conditional probability questions much simpler.

(iii) Independent Events

Level I

Question 1. A coin is tossed thrice and all 8 outcomes are equally likely. E : "The first throw results in head" F : "The last throw results in tail" Are the events independent ?
Answer:
The sample space \( S \) for tossing a coin three times contains 8 outcomes:
\( S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \)
Let us list the elements of events \( E \) and \( F \):
\( E = \{HHH, HHT, HTH, HTT\} \implies P(E) = \frac{4}{8} = \frac{1}{2} \)
\( F = \{HHT, HTT, THT, TTT\} \implies P(F) = \frac{4}{8} = \frac{1}{2} \)
The intersection \( E \cap F \) represents outcomes where the first is Head and the last is Tail:
\( E \cap F = \{HHT, HTT\} \implies P(E \cap F) = \frac{2}{8} = \frac{1}{4} \)
Let us check if \( P(E \cap F) = P(E) \cdot P(F) \):
\( P(E) \cdot P(F) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \)
Since \( P(E \cap F) = P(E) \cdot P(F) \), the events \( E \) and \( F \) are independent.
In simple words: Write out the outcomes for both events and find their intersection. Since the probability of the intersection (1/4) equals the product of their individual probabilities (1/2 times 1/2), the events are mathematically independent.

Exam Tip: To prove independence, always show that the test equation \( P(E \cap F) = P(E) \cdot P(F) \) holds true. This is the only way to earn full marks.

Question 2. Given P(A) = 1/4 , P(B) = 2/3 and P(A\(\cup\) B) = 3/4 . Are the events independent ?
Answer:
We know the addition formula for probability:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substituting the given values:
\( \frac{3}{4} = \frac{1}{4} + \frac{2}{3} - P(A \cap B) \)
\( P(A \cap B) = \frac{1}{4} + \frac{2}{3} - \frac{3}{4} = \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{6} = \frac{1}{6} \)
Now, let us calculate the product of the individual probabilities:
\( P(A) \cdot P(B) = \frac{1}{4} \cdot \frac{2}{3} = \frac{2}{12} = \frac{1}{6} \)
Since \( P(A \cap B) = P(A) \cdot P(B) = \frac{1}{6} \), the events \( A \) and \( B \) are independent.
In simple words: Find the intersection probability using the standard addition formula, which gives 1/6. Since this matches the product of individual probabilities (1/4 times 2/3), the events are independent.

Exam Tip: Be comfortable with fraction arithmetic, as many LPP and probability questions involve quick calculations using common denominators.

Question 3. If A and B are independent events, Find P(B) if P(A\(\cup\) B) = 0.60 and P(A) = 0.35.
Answer:
Since \( A \) and \( B \) are independent events, we have:
\( P(A \cap B) = P(A) \cdot P(B) \)
Let \( P(B) = p \). Using the addition theorem:
\( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Substituting the given values:
\( 0.60 = 0.35 + p - 0.35p \)
\( 0.60 - 0.35 = p(1 - 0.35) \)
\( 0.25 = 0.65p \)

\(\implies p = \frac{0.25}{0.65} = \frac{5}{13} \approx 0.385 \)
Thus, \( P(B) = \frac{5}{13} \).
In simple words: Use the addition formula and replace the intersection term with \( 0.35 \cdot P(B) \). Solve the resulting linear equation to find \( P(B) = 5/13 \).

Exam Tip: In algebraic probability questions, writing \( P(B) \) as a single variable like \( p \) makes the equation solving steps cleaner and easier to read.

(iv) Baye’s Theorem, Partition of Sample Space and Theorem of Total Probability

Level I

Question 1. A bag contains 6 red and 5 blue balls and another bag contains 5 red and 8 blue balls. A ball is drawn from the first bag and without noticing its colour is put in the second bag. A ball is drawn from the second bag . Find the probability that the ball drawn is blue in colour.
Answer:
Let \( E_1 \) be the event that the transferred ball is Red, and \( E_2 \) be the event that the transferred ball is Blue.
The probabilities of these events from the first bag are:
\( P(E_1) = \frac{6}{11} \)
\( P(E_2) = \frac{5}{11} \)
Let \( A \) be the event that the ball drawn from the second bag is Blue:
- If a Red ball is transferred (event \( E_1 \)), the second bag now contains 6 Red and 8 Blue balls (total 14):
\( P(A/E_1) = \frac{8}{14} = \frac{4}{7} \)
- If a Blue ball is transferred (event \( E_2 \)), the second bag now contains 5 Red and 9 Blue balls (total 14):
\( P(A/E_2) = \frac{9}{14} \)
Using the Theorem of Total Probability, the probability that the drawn ball is Blue is:
\( P(A) = P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) \)
\( P(A) = \left( \frac{6}{11} \cdot \frac{8}{14} \right) + \left( \frac{5}{11} \cdot \frac{9}{14} \right) \)
\( P(A) = \frac{48 + 45}{154} = \frac{93}{154} \)
Thus, the required probability is \( \frac{93}{154} \).
In simple words: The transferred ball can be either red or blue. Find the probability of drawing a blue ball in each of these two scenarios, multiply by the chance of the scenarios happening, and add them together.

Exam Tip: Clearly define the partition events \( E_1 \) and \( E_2 \) to set up a logical and easy-to-grade solution for the examiner.

Question 2. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both hearts . Find the probability of the lost card being a heart.
Answer:
Let \( E_1 \) be the event that the lost card is a Heart, and \( E_2 \) be the event that the lost card is not a Heart. The probabilities are:
\( P(E_1) = \frac{13}{52} = \frac{1}{4} \)
\( P(E_2) = \frac{39}{52} = \frac{3}{4} \)
Let \( A \) be the event that two Heart cards are drawn from the remaining 51 cards:
- If the lost card was a Heart (event \( E_1 \)), there are 12 Hearts left out of 51:
\( P(A/E_1) = \frac{C(12,2)}{C(51,2)} = \frac{12 \cdot 11}{51 \cdot 50} = \frac{132}{2550} \)
- If the lost card was not a Heart (event \( E_2 \)), there are 13 Hearts left out of 51:
\( P(A/E_2) = \frac{C(13,2)}{C(51,2)} = \frac{13 \cdot 12}{51 \cdot 50} = \frac{156}{2550} \)
Using Bayes' Theorem, the probability that the lost card is a Heart is:
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( P(E_1/A) = \frac{\frac{1}{4} \cdot \frac{132}{2550}}{\left( \frac{1}{4} \cdot \frac{132}{2550} \right) + \left( \frac{3}{4} \cdot \frac{156}{2550} \right)} \)
\( P(E_1/A) = \frac{132}{132 + 3 \cdot 156} = \frac{132}{132 + 468} = \frac{132}{600} = \frac{11}{50} \)
Therefore, the probability that the lost card was a Heart is \( \frac{11}{50} \).
In simple words: The lost card could be a heart or not a heart. Work out the probability of drawing two hearts in both situations. Use Bayes' theorem to reverse the logic and find that the chance of the lost card being a heart is 11/50.

Exam Tip: When using combination terms \( C(n,r) \), write out their expansions to demonstrate complete understanding of the counting steps.

Question 3. An insurance company insured 2000 scooter and 3000 motorcycles . The probability of an accident involving scooter is 0.01 and that of motorcycle is 0.02 . An insured vehicle met with an accident. Find the probability that the accidental vehicle was a motorcycle.
Answer:
Let \( E_1 \) be the event that the vehicle is a scooter, and \( E_2 \) be the event that the vehicle is a motorcycle.
The total number of insured vehicles is \( 2000 + 3000 = 5000 \).
\( P(E_1) = \frac{2000}{5000} = \frac{2}{5} \)
\( P(E_2) = \frac{3000}{5000} = \frac{3}{5} \)
Let \( A \) be the event that the selected vehicle met with an accident:
\( P(A/E_1) = 0.01 \)
\( P(A/E_2) = 0.02 \)
Using Bayes' Theorem, the probability that the vehicle was a motorcycle is:
\( P(E_2/A) = \frac{P(E_2) \cdot P(A/E_2)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( P(E_2/A) = \frac{\frac{3}{5} \cdot 0.02}{\left( \frac{2}{5} \cdot 0.01 \right) + \left( \frac{3}{5} \cdot 0.02 \right)} \)
\( P(E_2/A) = \frac{3 \cdot 0.02}{2 \cdot 0.01 + 3 \cdot 0.02} = \frac{0.06}{0.02 + 0.06} = \frac{0.06}{0.08} = \frac{3}{4} = 0.75 \)
Thus, the required probability is \( \frac{3}{4} \) (or 0.75).
In simple words: Find the fraction of scooters and motorcycles in the insured group. Multiply each by its accident rate, and apply Bayes' theorem to find that the chance the vehicle was a motorcycle is 3/4.

Exam Tip: Converting decimals like 0.01 and 0.02 can sometimes be simplified by leaving them as decimals and clearing the common factor later.

Question 4. A purse contains 2 silver and 4 copper coins. A second purse contains 4 silver and 3 copper coins. If a coin is pulled at random from one of the two purses, what is the probability that it is a silver coin.
Answer:
Let \( E_1 \) be the event of selecting the first purse, and \( E_2 \) be the event of selecting the second purse. Since purses are selected at random:
\( P(E_1) = P(E_2) = \frac{1}{2} \)
Let \( A \) be the event of drawing a silver coin:
- In the first purse, there are 2 silver and 4 copper coins (total 6):
\( P(A/E_1) = \frac{2}{6} = \frac{1}{3} \)
- In the second purse, there are 4 silver and 3 copper coins (total 7):
\( P(A/E_2) = \frac{4}{7} \)
Using the Theorem of Total Probability, the probability that the coin is silver is:
\( P(A) = P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) \)
\( P(A) = \left( \frac{1}{2} \cdot \frac{1}{3} \right) + \left( \frac{1}{2} \cdot \frac{4}{7} \right) \)
\( P(A) = \frac{1}{6} + \frac{2}{7} = \frac{7 + 12}{42} = \frac{19}{42} \)
Therefore, the probability of selecting a silver coin is \( \frac{19}{42} \).
In simple words: We have a 50% chance of choosing either purse. Find the chance of drawing a silver coin from the chosen purse in both scenarios and add them together.

Exam Tip: Be sure to keep the common denominator when adding fractions to avoid simple errors at the final step.

Question 5. Two thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting first class is 0.25 and that of a boy is getting a first class is 0.28. Find the probability that a student chosen at random will get first class marks in the subject.
Answer:
Let \( E_1 \) be the event of choosing a boy, and \( E_2 \) be the event of choosing a girl.
\( P(E_1) = \frac{2}{3} \)
\( P(E_2) = 1 - \frac{2}{3} = \frac{1}{3} \)
Let \( A \) be the event that the selected student gets first class marks:
\( P(A/E_1) = 0.28 \)
\( P(A/E_2) = 0.25 \)
Using the Theorem of Total Probability, the probability that the chosen student gets first class is:
\( P(A) = P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) \)
\( P(A) = \left( \frac{2}{3} \cdot 0.28 \right) + \left( \frac{1}{3} \cdot 0.25 \right) \)
\( P(A) = \frac{0.56 + 0.25}{3} = \frac{0.81}{3} = 0.27 \)
Thus, the required probability is \( 0.27 \).
In simple words: Find the weighted average of the first class rates based on the proportion of boys (2/3) and girls (1/3) in the class. This gives a final probability of 0.27.

Exam Tip: In total probability questions, write down the formula clearly before substituting numbers to secure steps marks.

Level II

Question 1. Find the probability of drawing a one-rupee coin from a purse with two compartments one of which contains 3 fifty-paise coins and 2 one-rupee coins and other contains 2 fifty-paise coins and 3 one-rupee coins.
Answer:
Let \( E_1 \) be the event of choosing the first compartment, and \( E_2 \) be the event of choosing the second compartment. Assuming compartments are chosen at random:
\( P(E_1) = P(E_2) = \frac{1}{2} \)
Let \( A \) be the event of drawing a one-rupee coin:
- The first compartment contains 3 fifty-paise and 2 one-rupee coins (total 5):
\( P(A/E_1) = \frac{2}{5} \)
- The second compartment contains 2 fifty-paise and 3 one-rupee coins (total 5):
\( P(A/E_2) = \frac{3}{5} \)
Using the Theorem of Total Probability, the probability of drawing a one-rupee coin is:
\( P(A) = P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) \)
\( P(A) = \left( \frac{1}{2} \cdot \frac{2}{5} \right) + \left( \frac{1}{2} \cdot \frac{3}{5} \right) = \frac{2 + 3}{10} = \frac{1}{2} \)
Therefore, the probability is \( \frac{1}{2} \).
In simple words: We have an equal chance of choosing either compartment. The average of drawing a one-rupee coin from both compartments is exactly 1/2.

Exam Tip: Always make sure to write out the partition probabilities \( P(E_1) = P(E_2) = \frac{1}{2} \) to show that you are accounting for the selection of the compartments.

Question 2. Suppose 5 men out of 100 and 25 women out of 1000 are good orator. An orator is chosen at random. Find the probability that a male person is selected. Assume that there are equal number of men and women.
Answer:
Let \( E_1 \) be the event of choosing a man, and \( E_2 \) be the event of choosing a woman. Since their numbers are equal:
\( P(E_1) = P(E_2) = \frac{1}{2} \)
Let \( A \) be the event that the chosen person is a good orator:
\( P(A/E_1) = \frac{5}{100} = 0.05 \)
\( P(A/E_2) = \frac{25}{1000} = 0.025 \)
Using Bayes' Theorem, the probability that the chosen orator is a man is:
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( P(E_1/A) = \frac{\frac{1}{2} \cdot 0.05}{\left( \frac{1}{2} \cdot 0.05 \right) + \left( \frac{1}{2} \cdot 0.025 \right)} \)
\( P(E_1/A) = \frac{0.05}{0.05 + 0.025} = \frac{0.05}{0.075} = \frac{2}{3} \)
Thus, the required probability is \( \frac{2}{3} \) (or approximately 0.67).
In simple words: Since there are equal numbers of men and women, find the fraction of good orators in both groups. Using Bayes' theorem, we find that the chance the chosen orator is a man is 2/3.

Exam Tip: Be sure to keep the fractions aligned so that the common factor \( \frac{1}{2} \) cancels out neatly in the numerator and denominator.

Question 3. A company has two plants to manufacture bicycles. The first plant manufactures 60 % of the bicycles and the second plant 40 % . Out of that 80 % of the bicycles are rated of standard quality at the first plant and 90 % of standard quality at the second plant. A bicycle is picked up at random and found to be standard quality. Find the probability that it comes from the second plant.
Answer:
Let \( E_1 \) be the event that the bicycle is made at the first plant, and \( E_2 \) be the event that the bicycle is made at the second plant.
\( P(E_1) = 60\% = 0.60 \)
\( P(E_2) = 40\% = 0.40 \)
Let \( A \) be the event that the selected bicycle is of standard quality:
\( P(A/E_1) = 80\% = 0.80 \)
\( P(A/E_2) = 90\% = 0.90 \)
Using Bayes' Theorem, the probability that the standard-quality bicycle comes from the second plant is:
\( P(E_2/A) = \frac{P(E_2) \cdot P(A/E_2)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( P(E_2/A) = \frac{0.40 \cdot 0.90}{(0.60 \cdot 0.80) + (0.40 \cdot 0.90)} \)
\( P(E_2/A) = \frac{0.36}{0.48 + 0.36} = \frac{0.36}{0.84} = \frac{3}{7} \)
Therefore, the probability is \( \frac{3}{7} \) (or approximately 0.43).
In simple words: Multiply each plant's market share by its standard quality rate. Using Bayes' theorem, we find that the chance the bicycle came from the second plant is 3/7.

Exam Tip: Leaving the percentages as decimals like 0.40 and 0.90 makes the multiplication much simpler and less prone to mistakes.

Level III

Question 1. A letter is known to have come either from LONDON or CLIFTON. On the envelope just has two consecutive letters ON are visible. What is the probability that the letter has come from (i) LONDON (ii) CLIFTON ?
Answer:
Let \( E_1 \) be the event that the letter came from LONDON, and \( E_2 \) be the event that the letter came from CLIFTON. Assuming equal prior probabilities:
\( P(E_1) = P(E_2) = \frac{1}{2} \)
Let \( A \) be the event that the two consecutive visible letters are "ON":
- For "LONDON", the possible consecutive pairs of letters are: LO, ON, ND, DO, ON. There are 5 pairs in total, and 2 of them are "ON":
\( P(A/E_1) = \frac{2}{5} \)
- For "CLIFTON", the possible consecutive pairs of letters are: CL, LI, IF, FT, TO, ON. There are 6 pairs in total, and 1 of them is "ON":
\( P(A/E_2) = \frac{1}{6} \)
Using Bayes' Theorem:

(i) Probability that the letter came from LONDON:
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( P(E_1/A) = \frac{\frac{1}{2} \cdot \frac{2}{5}}{\left( \frac{1}{2} \cdot \frac{2}{5} \right) + \left( \frac{1}{2} \cdot \frac{1}{6} \right)} = \frac{2/5}{2/5 + 1/6} = \frac{12}{17} \)

(ii) Probability that the letter came from CLIFTON:
\( P(E_2/A) = 1 - P(E_1/A) = 1 - \frac{12}{17} = \frac{5}{17} \)
Thus, the probability that the letter came from LONDON is \( \frac{12}{17} \), and from CLIFTON is \( \frac{5}{17} \).
In simple words: Write out the letter pairs for both words. Since "ON" appears twice in "LONDON" but only once in "CLIFTON", the probabilities are weighted towards LONDON, resulting in a 12/17 chance for LONDON and 5/17 for CLIFTON.

Exam Tip: Be sure to count the total consecutive pairs carefully. "LONDON" has 6 letters but only 5 consecutive pairs, while "CLIFTON" has 7 letters and 6 pairs.

Question 2. A test detection of a particular disease is not fool proof. The test will correctly detect the disease 90 % of the time, but will incorrectly detect the disease 1 % of the time. For a large population of which an estimated 0.2 % have the disease, a person is selected at random, given the test, and told that he has the disease. What are the chances that the person actually have the disease.
Answer:
Let \( E_1 \) be the event that the selected person has the disease, and \( E_2 \) be the event that the person does not have the disease.
\( P(E_1) = 0.2\% = 0.002 \)
\( P(E_2) = 1 - 0.002 = 0.998 \)
Let \( A \) be the event that the test result is positive:
- Correct detection (diseased person tests positive):
\( P(A/E_1) = 90\% = 0.90 \)
- False positive (healthy person tests positive):
\( P(A/E_2) = 1\% = 0.01 \)
Using Bayes' Theorem, the probability that the person actually has the disease given a positive test is:
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)} \)
\( P(E_1/A) = \frac{0.002 \cdot 0.90}{(0.002 \cdot 0.90) + (0.998 \cdot 0.01)} \)
\( P(E_1/A) = \frac{0.0018}{0.0018 + 0.00998} = \frac{0.0018}{0.01178} = \frac{90}{589} \approx 0.153 \)
Thus, the probability that the person actually has the disease is \( \frac{90}{589} \) (or approximately 15.3%).
In simple words: Multiply the tiny fraction of people with the disease by the test's accuracy. Using Bayes' theorem, we find that even with a positive test, there is only about a 15.3% chance of actually having the disease due to false positives in the healthy population.

Exam Tip: This counter-intuitive result (low probability of having the disease despite a positive test) is a classic example of Bayes' Theorem applied to screening tests and is very common in board exams.

Question 3. Given three identical boxes I, II and III each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in box III , there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold ? [CBSE 2011]
Answer:
Let \( E_1, E_2, E_3 \) be the events of choosing Box I, Box II, and Box III respectively.
\( P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \)
Let \( A \) be the event that the drawn coin is gold:
- Box I has 2 gold coins:
\( P(A/E_1) = 1 \)
- Box II has 0 gold coins:
\( P(A/E_2) = 0 \)
- Box III has 1 gold and 1 silver coin:
\( P(A/E_3) = \frac{1}{2} \)
We want to find the probability that the other coin is gold, which means the chosen box is Box I. We need to find \( P(E_1/A) \):
\( P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2) + P(E_3) \cdot P(A/E_3)} \)
\( P(E_1/A) = \frac{\frac{1}{3} \cdot 1}{\left( \frac{1}{3} \cdot 1 \right) + \left( \frac{1}{3} \cdot 0 \right) + \left( \frac{1}{3} \cdot \frac{1}{2} \right)} = \frac{1}{1 + 0 + 1/2} = \frac{2}{3} \)
Therefore, the required probability is \( \frac{2}{3} \).
In simple words: The chosen coin is gold. This means we must have selected either Box I or Box III. Using Bayes' theorem, we find there is a 2/3 chance that we chose Box I (where both coins are gold).

Exam Tip: This is a highly repeated CBSE question. Writing the formula for Bayes' Theorem with three partition events clearly is crucial for earning full marks.

(v) Random Variables & Probability Distribution Mean & Variance of Random Variables

Level I

Question 1. Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of spades
Answer:
Let \( X \) be the random variable denoting the number of spades drawn in two draws. The values \( X \) can take are \( 0, 1, 2 \).
Total cards = 52. Spades = 13. Non-spades = 39.
Probability of drawing a spade on any draw, \( p = \frac{13}{52} = \frac{1}{4} \).
Probability of drawing a non-spade on any draw, \( q = 1 - p = \frac{3}{4} \).
Since the cards are drawn with replacement, the draws are independent. We calculate the probabilities:
- \( P(X = 0) = q^2 = \left( \frac{3}{4} \right)^2 = \frac{9}{16} \)
- \( P(X = 1) = 2pq = 2 \left( \frac{1}{4} \right) \left( \frac{3}{4} \right) = \frac{6}{16} = \frac{3}{8} \)
- \( P(X = 2) = p^2 = \left( \frac{1}{4} \right)^2 = \frac{1}{16} \)
The probability distribution of \( X \) is:

In simple words: Since we replace the card, the draws are independent. The chance of getting no spades is 9/16, one spade is 6/16 (or 3/8), and two spades is 1/16.

Exam Tip: Always make sure to write the probability distribution in a tabular format as shown above to ensure your answer is complete.

Question 2. 4 defective apples are accidentally mixed with 16 good ones. Three apples are drawn at random from the mixed lot. Find the probability distribution of the number of defective apples.
Answer:
Let \( X \) be the number of defective apples drawn. The values \( X \) can take are \( 0, 1, 2, 3 \).
Total apples = \( 4 + 16 = 20 \). Defective apples = 4. Good apples = 16.
The total number of ways to draw 3 apples from 20 is \( C(20,3) = \frac{20 \cdot 19 \cdot 18}{6} = 1140 \).
We calculate the probabilities for each value of \( X \):
- \( P(X = 0) = \frac{C(16,3) \cdot C(4,0)}{C(20,3)} = \frac{560}{1140} = \frac{28}{57} \)
- \( P(X = 1) = \frac{C(16,2) \cdot C(4,1)}{C(20,3)} = \frac{120 \cdot 4}{1140} = \frac{24}{57} \)
- \( P(X = 2) = \frac{C(16,1) \cdot C(4,2)}{C(20,3)} = \frac{16 \cdot 6}{1140} = \frac{8}{95} \)
- \( P(X = 3) = \frac{C(16,0) \cdot C(4,3)}{C(20,3)} = \frac{4}{1140} = \frac{1}{285} \)
The probability distribution of \( X \) is:

In simple words: Since we are drawing apples without replacement, we use combination formulas. Calculate the ways to choose defective and good apples, then divide by the total ways to choose 3 apples from 20.

Exam Tip: Verify that the sum of the probabilities in your distribution table equals 1 to ensure you have not made any arithmetic errors.

Question 3. A random variable X is specified by the following distribution Find the variance of the distribution. X: 2, 3, 4 P(X): 0.3, 0.4, 0.3
Answer:
First, we find the mean \( \mu = E(X) \) of the distribution:
\( E(X) = \sum X \cdot P(X) = 2(0.3) + 3(0.4) + 4(0.3) = 0.6 + 1.2 + 1.2 = 3.0 \)
Now, find \( E(X^2) \):
\( E(X^2) = \sum X^2 \cdot P(X) = 2^2(0.3) + 3^2(0.4) + 4^2(0.3) = 4(0.3) + 9(0.4) + 16(0.3) \)
\( E(X^2) = 1.2 + 3.6 + 4.8 = 9.6 \)
The variance is given by:
\( \sigma^2 = E(X^2) - [E(X)]^2 = 9.6 - 3.0^2 = 9.6 - 9.0 = 0.6 \)
Therefore, the variance of the distribution is \( 0.6 \).
In simple words: Calculate the mean by multiplying each number by its probability. Then do the same for the squared values, and use the variance formula to get the final answer of 0.6.

Exam Tip: Be sure to write out the formula for variance \( \sigma^2 = E(X^2) - [E(X)]^2 \) clearly before substituting values.

Level III

Question 1. A coin is biased so that the head is 3 times as likely to occur as a tail. If the coin is tossed twice.Find the probability distribution of the number of tails.
Answer:
Let the probability of a tail be \( P(T) = p \). Since the head is 3 times as likely as the tail, the probability of a head is \( P(H) = 3p \).
Since the sum of the probabilities is 1:
\( p + 3p = 1 \implies 4p = 1 \implies p = \frac{1}{4} \)
Thus, \( P(T) = \frac{1}{4} \) and \( P(H) = \frac{3}{4} \).
Let \( X \) be the random variable denoting the number of tails in two tosses. \( X \) can take values \( 0, 1, 2 \):
- \( P(X = 0) = P(H) \cdot P(H) = \left( \frac{3}{4} \right)^2 = \frac{9}{16} \)
- \( P(X = 1) = P(H) \cdot P(T) + P(T) \cdot P(H) = 2 \left( \frac{3}{4} \right) \left( \frac{1}{4} \right) = \frac{6}{16} \)
- \( P(X = 2) = P(T) \cdot P(T) = \left( \frac{1}{4} \right)^2 = \frac{1}{16} \)
The probability distribution of \( X \) is:

In simple words: The coin is biased so head has a 3/4 chance and tail has a 1/4 chance. Multiply the individual probabilities to get the distribution for getting 0, 1, or 2 tails.

Exam Tip: First find the individual probabilities \( p \) and \( q \) using the ratio relation before attempting to calculate the binomial terms.

Question 2. The sum of mean and variance of a binomial distribution for 5 trials be 1.8. Find the probability distribution.
Answer:
For a binomial distribution with \( n = 5 \) trials, the mean is \( np = 5p \) and the variance is \( npq = 5pq = 5p(1-p) \).
Given:
\( \text{Mean} + \text{Variance} = 1.8 \)
\( 5p + 5p(1-p) = 1.8 \)
\( 5p + 5p - 5p^2 = 1.8 \)
\( 10p - 5p^2 = 1.8 \)
Multiply by 10 to clear decimals:
\( 50p^2 - 100p + 18 = 0 \)
Dividing by 2:
\( 25p^2 - 50p + 9 = 0 \)
Factoring the quadratic equation:
\( 25p^2 - 45p - 5p + 9 = 0 \)
\( 5p(5p - 9) - 1(5p - 9) = 0 \)
\( (5p - 1)(5p - 9) = 0 \)
Since \( p \le 1 \), the value \( p = \frac{9}{5} \) is rejected. Therefore:
\( p = \frac{1}{5} = 0.2 \implies q = 0.8 \)
The probability distribution of binomial distribution is given by:
\( P(X = x) = C(5,x) \cdot (0.8)^{5-x} \cdot (0.2)^x \quad \text{for } x = 0, 1, 2, 3, 4, 5 \).
In simple words: Write out the mean and variance as formulas. Solve the resulting quadratic equation to find \( p = 0.2 \), and then write down the general binomial term.

Exam Tip: Always make sure to state that the probability value \( p \) cannot be greater than 1, which mathematically justifies discarding the other root of the quadratic equation.

Question 3. The mean and variance of a binomial distribution are 4/3 and 8/9 respectively. Find P(X \(\ge\) 1).
Answer:
For a binomial distribution, the mean is \( np = \frac{4}{3} \) and the variance is \( npq = \frac{8}{9} \).
Dividing the variance by the mean gives:
\( q = \frac{npq}{np} = \frac{8/9}{4/3} = \frac{8}{9} \cdot \frac{3}{4} = \frac{2}{3} \)
So, the probability of success is:
\( p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3} \)
Now, using the mean to find \( n \):
\( n \cdot \frac{1}{3} = \frac{4}{3} \implies n = 4 \)
We need to find \( P(X \ge 1) \):
\( P(X \ge 1) = 1 - P(X = 0) = 1 - C(4,0) \cdot q^4 \)
\( P(X \ge 1) = 1 - \left( \frac{2}{3} \right)^4 = 1 - \frac{16}{81} = \frac{65}{81} \)
Thus, the required probability is \( \frac{65}{81} \).
In simple words: Divide variance by mean to find \( q = 2/3 \) and \( p = 1/3 \). This tells us there are 4 trials. The probability of getting at least one success is 1 minus the chance of getting no successes, which is 65/81.

Exam Tip: Dividing variance by mean is a standard technique to find the probability of failure \( q \) instantly in binomial questions.

(vi) Bernoulli,s Trials and Binomial Distribution

Level II

Question 1. If a die is thrown 5 times, what is the chance that an even number will come up exactly 3 times.
Answer:
Let throwing a die be a Bernoulli trial. The number of trials is \( n = 5 \).
Success is defined as getting an even number \( \{2, 4, 6\} \). The probability of success is:
\( p = \frac{3}{6} = \frac{1}{2} \implies q = \frac{1}{2} \)
The probability of getting exactly 3 successes is:
\( P(X = 3) = C(5,3) \cdot p^3 \cdot q^2 \)
\( P(X = 3) = 10 \cdot \left( \frac{1}{2} \right)^3 \cdot \left( \frac{1}{2} \right)^2 = \frac{10}{32} = \frac{5}{16} \)
Thus, the probability is \( \frac{5}{16} \).
In simple words: The chance of getting an even number is 1/2. We want exactly 3 even numbers in 5 rolls, which we calculate using the binomial formula to get 5/16.

Exam Tip: Be sure to write the formula for binomial distribution \( P(X=r) = C(n,r) p^r q^{n-r} \) before substituting numbers to secure steps marks.

Question 2. An experiment succeeds twice as often it fails. Find the probability that in the next six trials, there will be at least 4 success.
Answer:
The probability of success \( p \) is twice that of failure \( q \):
\( p = 2q \)
Since \( p + q = 1 \):
\( 2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3} \implies p = \frac{2}{3} \)
The number of trials is \( n = 6 \). We need to find the probability of getting at least 4 successes:
\( P(X \ge 4) = P(X = 4) + P(X = 5) + P(X = 6) \)
Calculating each term:
- \( P(X = 4) = C(6,4) \cdot \left( \frac{2}{3} \right)^4 \cdot \left( \frac{1}{3} \right)^2 = 15 \cdot \frac{16}{729} = \frac{240}{729} \)
- \( P(X = 5) = C(6,5) \cdot \left( \frac{2}{3} \right)^5 \cdot \left( \frac{1}{3} \right)^1 = 6 \cdot \frac{32}{729} = \frac{192}{729} \)
- \( P(X = 6) = C(6,6) \cdot \left( \frac{2}{3} \right)^6 = 1 \cdot \frac{64}{729} = \frac{64}{729} \)
Total probability is:
\( P(X \ge 4) = \frac{240 + 192 + 64}{729} = \frac{496}{729} \)
Thus, the required probability is \( \frac{496}{729} \).
In simple words: The success rate is 2/3. Calculate the chance of getting exactly 4, 5, and 6 successes out of 6 trials, and add them up to find the final probability of 496/729.

Exam Tip: When evaluating multiple terms, keep the common denominator \( 3^6 = 729 \) in all calculations to make the final addition simple.

Question 3. A pair of dice is thrown 200 times. If getting a sum 9 is considered a success, find the mean and variance of the number of success.
Answer:
Let rolling a pair of dice be a Bernoulli trial. The number of trials is \( n = 200 \).
A success is defined as getting a sum of 9. The successful outcomes are:
\( \{(3,6), (4,5), (5,4), (6,3)\} \)
The number of favorable outcomes is 4. The total outcomes in rolling two dice is 36.
The probability of success is:
\( p = \frac{4}{36} = \frac{1}{9} \implies q = 1 - p = \frac{8}{9} \)
For a binomial distribution:
- Mean = \( np = 200 \cdot \frac{1}{9} = \frac{200}{9} \approx 22.22 \)
- Variance = \( npq = 200 \cdot \frac{1}{9} \cdot \frac{8}{9} = \frac{1600}{81} \approx 19.75 \)
Therefore, the mean is \( \frac{200}{9} \) and the variance is \( \frac{1600}{81} \).
In simple words: The chance of rolling a sum of 9 with two dice is 1/9. In 200 rolls, we expect a mean of about 22.22 successes, with a variance of about 19.75.

Exam Tip: Be sure to find the probability of success \( p \) by listing the favorable coordinate pairs before calculating the mean and variance.

Questions for Self Evaluation

Question 1. A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Find the probability that the number is divisible by 5.
Answer:
The digits available are \( \{1, 2, 3, 5\} \).
The total number of 4-digit numbers that can be formed without repetition is:
\( n(\text{Total}) = 4! = 24 \)
For a number to be divisible by 5, the last digit must be 5. If we fix 5 at the units place, the remaining 3 places can be filled with the remaining 3 digits \( \{1, 2, 3\} \) in:
\( n(\text{Favorable}) = 3! = 6 \text{ ways} \)
The required probability is:
\( P = \frac{n(\text{Favorable})}{n(\text{Total})} = \frac{6}{24} = \frac{1}{4} = 0.25 \)
Therefore, the probability is \( 0.25 \).
In simple words: There are 24 total ways to arrange the four digits. For the number to be divisible by 5, the units place must be 5, which leaves 6 successful combinations. The probability is 6/24, or 1/4.

Exam Tip: Simple permutation rules \( n! \) can be used to quickly find both the total and favorable outcomes in digit-arrangement questions.

Question 2. The probability that an event happens in one trial of an experiment is 0.4. Three independent trials of an experiment are performed. Find the probability that the event happens at least once.
Answer:
The probability of the event happening in one trial is \( p = 0.4 \).
The probability of the event not happening in one trial is:
\( q = 1 - p = 0.6 \)
The number of trials is \( n = 3 \). We need to find the probability that the event happens at least once:
\( P(X \ge 1) = 1 - P(X = 0) \)
\( P(X \ge 1) = 1 - q^3 = 1 - (0.6)^3 = 1 - 0.216 = 0.784 \)
Thus, the required probability is \( 0.784 \).
In simple words: The chance of the event not happening in any of the three trials is \( 0.6 \cdot 0.6 \cdot 0.6 = 0.216 \). The chance of it happening at least once is the opposite, which is 1 minus 0.216, giving 0.784.

Exam Tip: "At least once" problems are almost always solved much faster by subtracting the "none" probability from 1.

Question 3. A football match is either won, draw or lost by the host country’s team. So there are three ways of forecasting the result of any one match, one correct and two incorrect. Find the probability of forecasting at least three correct results for four matches.
Answer:
Let guessing the correct result of a match be a success. Since there are 3 possible results and only 1 is correct:
\( p = \frac{1}{3} \implies q = \frac{2}{3} \)
The number of matches is \( n = 4 \). We need to find the probability of getting at least 3 correct forecasts:
\( P(X \ge 3) = P(X = 3) + P(X = 4) \)
- \( P(X = 3) = C(4,3) \cdot p^3 \cdot q^1 = 4 \cdot \left( \frac{1}{3} \right)^3 \cdot \left( \frac{2}{3} \right)^1 = \frac{8}{81} \)
- \( P(X = 4) = C(4,4) \cdot p^4 = 1 \cdot \left( \frac{1}{3} \right)^4 = \frac{1}{81} \)
The total probability is:
\( P(X \ge 3) = \frac{8 + 1}{81} = \frac{9}{81} = \frac{1}{9} \)
Therefore, the probability is \( \frac{1}{9} \).
In simple words: The chance of guessing correctly is 1/3. Use the binomial formula to find the probability of getting exactly 3 or 4 correct guesses out of 4 matches, and add them up to find the final probability of 1/9.

Exam Tip: Be careful with the calculations; keeping the denominator as \( 3^4 = 81 \) simplifies the addition of the two probability terms.

Question 4. A candidate has to reach the examination center in time. Probability of him going by bus ore scooter or by other means of transport are 3/10 , 1/10 , 3/5 respectively. The probability that he will be late is 1/4 and 1/3 respectively. But he reaches in time if he uses other mode of transport. He reached late at the centre. Find the probability that he traveled by bus.
Answer:
Let \( E_1 \) be the event that the candidate goes by bus, \( E_2 \) be the event that he goes by scooter, and \( E_3 \) be the event that he goes by other modes.
\( P(E_1) = \frac{3}{10} \)
\( P(E_2) = \frac{1}{10} \)
\( P(E_3) = \frac{3}{5} = \frac{6}{10} \)
Let \( L \) be the event that the candidate is late:
- Probability of being late by bus:
\( P(L/E_1) = \frac{1}{4} \)
- Probability of being late by scooter:
\( P(L/E_2) = \frac{1}{3} \)
- Probability of being late by other mode (he reaches in time):
\( P(L/E_3) = 0 \)
Using Bayes' Theorem, the probability that he traveled by bus is:
\( P(E_1/L) = \frac{P(E_1) \cdot P(L/E_1)}{P(E_1) \cdot P(L/E_1) + P(E_2) \cdot P(L/E_2) + P(E_3) \cdot P(L/E_3)} \)
\( P(E_1/L) = \frac{\frac{3}{10} \cdot \frac{1}{4}}{\left( \frac{3}{10} \cdot \frac{1}{4} \right) + \left( \frac{1}{10} \cdot \frac{1}{3} \right) + \left( \frac{6}{10} \cdot 0 \right)} \)
\( P(E_1/L) = \frac{3/40}{3/40 + 1/30} = \frac{3/40}{13/120} = \frac{3}{40} \cdot \frac{120}{13} = \frac{9}{13} \)
Thus, the required probability is \( \frac{9}{13} \).
In simple words: Use the given probabilities to find the total rate of being late. Since he is never late using other modes, use Bayes' theorem to find that the chance he took the bus given he is late is 9/13.

Exam Tip: Be sure to include the term with probability 0 in your formula to show that you are fully applying the partition of the sample space.

Question 5. Let X denote the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that P(X = x) = {kx, if x = 0, or 1; 2kx, if x = 2; k(5 - x), if x = 3 or 4; , k is a + ve constant. Find the mean and variance of the probability distribution. 1
Answer:
First, we find the value of \( k \) by setting the sum of all probabilities in the distribution to 1:
- \( P(X = 0) = k(0) = 0 \)
- \( P(X = 1) = k(1) = k \)
- \( P(X = 2) = 2k(2) = 4k \)
- \( P(X = 3) = k(5 - 3) = 2k \)
- \( P(X = 4) = k(5 - 4) = k \)
Since the sum of the probabilities must be 1:
\( \sum P(X) = 0 + k + 4k + 2k + k = 1 \)
\( 8k = 1 \implies k = \frac{1}{8} \)
The probability distribution of \( X \) is:

Now, find the mean \( \mu = E(X) \):
\( E(X) = \sum X \cdot P(X) = 0(0) + 1\left(\frac{1}{8}\right) + 2\left(\frac{4}{8}\right) + 3\left(\frac{2}{8}\right) + 4\left(\frac{1}{8}\right) = \frac{1 + 8 + 6 + 4}{8} = \frac{19}{8} = 2.375 \)
Next, find \( E(X^2) \):
\( E(X^2) = \sum X^2 \cdot P(X) = 0^2(0) + 1^2\left(\frac{1}{8}\right) + 2^2\left(\frac{4}{8}\right) + 3^2\left(\frac{2}{8}\right) + 4^2\left(\frac{1}{8}\right) = \frac{1 + 16 + 18 + 16}{8} = \frac{51}{8} = 6.375 \)
The variance \( \sigma^2 \) is given by:
\( \sigma^2 = E(X^2) - [E(X)]^2 = \frac{51}{8} - \left( \frac{19}{8} \right)^2 = \frac{51}{8} - \frac{361}{64} = \frac{408 - 361}{64} = \frac{47}{64} \approx 0.734 \)
Therefore, the mean is \( 2.375 \) and the variance is \( \frac{47}{64} \) (or approximately 0.734).
In simple words: Find \( k = 1/8 \) by summing up all the pieces. Use this value to compute the average (mean) as 2.375 and the spread (variance) as 47/64.

Exam Tip: Always make sure to write out the probability table with actual values after finding the constant \( k \) before calculating the mean and variance.

Question 6. A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
Answer:
For the third six to occur on the sixth throw:
1. Exactly two sixes must be obtained in the first 5 throws.
2. The sixth throw must result in a six.
Let a success be getting a six on a single roll:
\( p = \frac{1}{6} \implies q = \frac{5}{6} \)
The probability of getting exactly 2 sixes in 5 throws is:
\( P_1 = C(5,2) \cdot p^2 \cdot q^3 = 10 \cdot \left( \frac{1}{6} \right)^2 \cdot \left( \frac{5}{6} \right)^3 = 10 \cdot \frac{125}{7776} = \frac{1250}{7776} \)
The probability of getting a six on the sixth throw is:
\( P_2 = \frac{1}{6} \)
Since the trials are independent, the required probability is the product:
\( P = P_1 \cdot P_2 = \frac{1250}{7776} \cdot \frac{1}{6} = \frac{625}{23328} \)
Thus, the required probability is \( \frac{625}{23328} \).
In simple words: Find the chance of getting exactly 2 sixes in the first 5 rolls (1250/7776), and multiply it by the chance of rolling a six on the last throw (1/6) to get 625/23328.

Exam Tip: This is a classic sequential binomial question. The key is to fix the last success at the designated final trial and apply the binomial distribution to the preceding trials.

Question 7. On a multiple choice examination with three possible answers(out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ?
Answer:
Let guessing the correct answer to a question be a success. The number of trials is \( n = 5 \).
Since there are 3 choices and only 1 is correct:
\( p = \frac{1}{3} \implies q = \frac{2}{3} \)
We need to find the probability of getting 4 or more correct answers:
\( P(X \ge 4) = P(X = 4) + P(X = 5) \)
- \( P(X = 4) = C(5,4) \cdot p^4 \cdot q^1 = 5 \cdot \left( \frac{1}{3} \right)^4 \cdot \left( \frac{2}{3} \right)^1 = \frac{10}{243} \)
- \( P(X = 5) = C(5,5) \cdot p^5 = 1 \cdot \left( \frac{1}{3} \right)^5 = \frac{1}{243} \)
The total probability is:
\( P(X \ge 4) = \frac{10 + 1}{243} = \frac{11}{243} \)
Thus, the required probability is \( \frac{11}{243} \).
In simple words: The success rate is 1/3. Calculate the chance of getting exactly 4 or 5 correct answers out of 5 questions, and add them up to find the final probability of 11/243.

Exam Tip: Be careful with the base power calculations; keeping the denominator as \( 3^5 = 243 \) makes adding the final terms simple.

Question 8. Two cards are drawn simultaneously (or successively) from a well shuffled pack of 52 cards. Find the mean and variance of the number of red cards.
Answer:
Let \( X \) be the random variable representing the number of red cards drawn. \( X \) can take values \( 0, 1, 2 \).
Total cards = 52. Red cards = 26. Black cards = 26.
Since the drawing is simultaneous (which is equivalent to drawing without replacement):
- \( P(X = 0) = \frac{C(26,0) \cdot C(26,2)}{C(52,2)} = \frac{26 \cdot 25}{52 \cdot 51} = \frac{25}{102} \)
- \( P(X = 1) = \frac{C(26,1) \cdot C(26,1)}{C(52,2)} = \frac{26 \cdot 26}{26 \cdot 51} = \frac{52}{102} \)
- \( P(X = 2) = \frac{C(26,2) \cdot C(26,0)}{C(52,2)} = \frac{26 \cdot 25}{52 \cdot 51} = \frac{25}{102} \)
The mean \( E(X) \) of the distribution is:
\( E(X) = \sum X \cdot P(X) = 0\left( \frac{25}{102} \right) + 1\left( \frac{52}{102} \right) + 2\left( \frac{25}{102} \right) = \frac{102}{102} = 1 \)
To find the variance, first find \( E(X^2) \):
\( E(X^2) = \sum X^2 \cdot P(X) = 0^2\left( \frac{25}{102} \right) + 1^2\left( \frac{52}{102} \right) + 2^2\left( \frac{25}{102} \right) = \frac{52 + 100}{102} = \frac{152}{102} = \frac{76}{51} \)
The variance is:
\( \sigma^2 = E(X^2) - [E(X)]^2 = \frac{76}{51} - 1^2 = \frac{25}{51} \approx 0.49 \)
Therefore, the mean is \( 1 \) and the variance is \( \frac{25}{51} \) (or approximately 0.49).
In simple words: Since we draw without replacement, use combination counting to set up the probability distribution. The average (mean) is 1, and the variance (spread) is 25/51.

Exam Tip: Notice that drawing simultaneously is equivalent to non-replacement, which means you must use combinations rather than the binomial formula.