CBSE Class 10 Maths Probability Worksheet Set 02

Question. A letter is chosen at random from the word ‘ASSASSINATION’. The probability that it is a vowel is 
(a) \( \frac{6}{13} \)
(b) \( \frac{7}{13} \)
(c) \( \frac{31}{3} \)
(d) \( \frac{3}{13} \)
Answer: (a) \( \frac{6}{13} \)

Question. A number ‘x’ is chosen at random from the numbers -4, -3, -2, -1, 0, 1, 2, 3, 4, 5. The probability that \( \lvert x \rvert < 3 \) is  
(a) 1
(b) 0
(c) \( \frac{1}{2} \)
(d) \( \frac{7}{10} \)
Answer: (c) \( \frac{1}{2} \)

Question. An unbiased die is thrown once. The probability of getting an odd number is 
(a) \( \frac{1}{3} \)
(b) \( \frac{1}{2} \)
(c) \( \frac{2}{5} \)
(d) \( \frac{2}{3} \)
Answer: (b) \( \frac{1}{2} \)

Question. A number is selected at random from 1 to 75. The probability that it is a perfect square is 
(a) \( \frac{10}{75} \)
(b) \( \frac{8}{75} \)
(c) \( \frac{6}{75} \)
(d) \( \frac{4}{75} \)
Answer: (b) \( \frac{8}{75} \)

Question. A lot consists of 40 mobile phones of which 32 are good, 3 have only minor defects and 5 have major defects. Ram will buy a phone if it is good or have minor defects. One phone is selected at random. The probability that it is acceptable to Ram is___.  
(a) \( \frac{3}{40} \)
(b) \( \frac{4}{5} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{7}{8} \)
Answer: (d) \( \frac{7}{8} \)

Question. A black dice and a white dice are thrown at the same time. Write all the possible outcomes. What is the probability that the difference of the numbers appearing on the top of the two dice is 2?  
Answer: Consider the set of ordered pairs
{(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)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}
Clearly, there are 36 elementary events.
\( \therefore \) n(Total number of throws) = 36
number of pairs such that difference of the numbers appearing on top of the two dice is 2 can be selected as listed below:
{(1,3)(2,4)(3,1)(3,5)(4,2)(4,6)(5,3)(6,4)}
Therefore, n(Favourable events) = 8
P(difference of the number is 2) = (Number of pairs such that difference of the numbers Appearing on the top of the two dice is 2) / (Total number of throws)
= \( \frac{8}{36} = \frac{2}{9} \)

Question. Three unbiased coins are tossed together. Find the probability of getting at least two heads?  
Answer: If any of the elementary events HHH, HHT, HTH, and THH is an outcome, then we say that the event "Getting at least two heads" occurs.
Favourable number of elementary events = 4
total no. of possible events when three coins are tossed = 8
Hence, required probability = \( \frac{4}{8} = \frac{1}{2} \)

Question. Why is tossing a coin considered as the way of deciding which team should get the ball at the beginning of a football match? 
Answer: Proabibilty of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} \)
Probability of head = P(H) = \( \frac{1}{2} \)
Probability of tail = P(T) = \( \frac{1}{2} \)
i.e. \( P(H) = P(T) = \frac{1}{2} \)
Probability of getting head and tail both are same.
\( \therefore \) Tossing a coin considered to be fairway.

Question. Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one dice and a multiple of 3 on the other.  
Answer: Two dice are thrown simultaneously. We have to find the probability of getting a multiple of 2 on one dice and a multiple of 3 on the other.
Let A be the event of getting a multiple of 2 on one die and a multiple of 3 on the other.
Then, the elementary events favourable to A are:
(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4).
Favourable number of elementary events = 11
Hence, required probability = \( \frac{11}{36} \)

Question. In a simultaneous throw of a pair of dice, find the probability of getting a number other than 5 on any dice. 
Answer: Favourable outcomes of a number other than 5 on any dice =
{(1,1)(1,2)(1,3)(1,4)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,6)
(6,1)(6,2)(6,3)(6,4)(6,6)}
Therefore, favourable number of cases to the event = 25
\( \therefore \) Probability of a number other than 5 on any dice = \( \frac{\text{number of favourable outcomes}}{\text{number of total outcomes}} = \frac{25}{36} \)

Question. Two coins are tossed together. Find the probability of getting both heads or both tails. 
Answer: Two coins are tossed together.
Possibilities are HH, HT, TH, TT
Total outcomes = 4
Both heads or both tails = HH, TT
Number of favourable outcome = 2
\( \text{Probability} = \frac{\text{Number of favorable outcome}}{\text{Total number of outcome}} \)
P(HH or TT) = \( \frac{2}{4} = \frac{1}{2} \)

Question. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is(i) red (ii) black or white (iii) not black. 
Answer: Proabibilty of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} \)
Total number of balls = 5 + 7 + 3 = 15
(i) Number of red balls = 7
\( \therefore \) P(drawing a red ball) = \( \frac{7}{15} \)
(ii) Number of black or white balls = 5 + 3 = 8
\( \therefore \) P(drawing a black or white ball) = \( \frac{8}{15} \)
(iii) Number of balls which are not black = 15 - 5 = 10
\( \therefore \) P(drawing a ball that is not black) = \( \frac{10}{15} = \frac{2}{3} \)
Hence, the probability of getting a red ball, a black or white ball and a not black ball are \( \frac{7}{15} \), \( \frac{8}{15} \) and \( \frac{2}{3} \) respectively.

Question. A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out?  
i. an orange flavoured candy?
ii. a lemon flavoured candy?
Answer: (i) The probability that she takes out an orange flavoured candy is 0 because the bag contains lemon flavoured candies only.
(ii) Probability that she takes out a lemon flavoured candy is 1 because the bag contains lemon flavoured candies only.

Question. Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken out card is  
i. a prime number less than 10
ii. a number which is a perfect square.
Answer: According to question we are given that Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random.
Therefore All possible outcomes are 5, 6, 7, 8 ............... 50.
Number of all possible outcomes = 46
(i) Out of the given numbers, the prime numbers less than 10 are 5 and 7.
Suppose \( E_1 \) be the event of getting a prime number less than 10.
Then, number of favorable outcomes = 2
Therefore, P(getting a prime number less than 10) = P(E) = \( \frac{2}{46} = \frac{1}{23} \)
(ii) Out of the given numbers, the perfect squares are 9, 16, 25, 36 and 49.
Suppose \( E_2 \) be the event of getting a perfect square.
Then, number of favorable outcomes = 5
Therefore, P(getting a perfect square) = P(E) = \( \frac{5}{46} \)

Question. Cards numbered 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is (i) an odd number, (ii) a perfect square number, (iii) divisible by 5, (iv) a prime number less than 20.  
Answer: According to the question,
All possible outcomes are 11, 12, 13, . . . , 60.
Total number of possible outcomes = (60 – 10) = 50.
(i) Suppose, \( E_1 \) be the event that the number on the drawn card is an odd number.
\( \Rightarrow \) the favourable outcomes are 11, 13, 15, . . . , 59.
Clearly, these numbers form an AP with a = 11 and d = 2.
\( T_n = 59 \Rightarrow 11 + (n - 1) \times 2 = 59 \Rightarrow (n - 1) \times 2 = 48 \Rightarrow n - 1 = 24 \Rightarrow n = 25 \)
So, the number of favourable outcomes = 25.
\( \therefore \) P(getting an odd number) = \( P(E_1) = \frac{25}{50} = \frac{1}{2} \).
(ii) Suppose, \( E_2 \) be the event that the number on the drawn card is a perfect square number.
\( \Rightarrow \) the favourable outcomes are 16, 25, 36, 49.
The number of favourable outcomes = 4.
\( \therefore \) P(getting a perfect square number) \( P(E_2) = \frac{4}{50} = \frac{2}{25} \).
(iii) Suppose, \( E_3 \) be the event that the number on the drawn card is divisible by 5.
\( \Rightarrow \) the favourable outcomes are 15, 20, 25, . . . , 60.
Clearly, these numbers form an AP with a = 15 and d = 5.
\( T_m = 60 \Rightarrow 15 + (m - 1) \times 5 = 60 \Rightarrow (m - 1) \times 5 = 45 \Rightarrow m - 1 = 9 \Rightarrow m = 10 \)
So, the number of favourable outcomes = 10.
P(getting a number divisible by 5) = \( P(E_3) = \frac{10}{50} = \frac{1}{5} \)
(iv) Suppose, \( E_4 \) be the event that the number on the drawn card is a prime number less than 20.
\( \Rightarrow \) the favourable outcomes are 11, 13, 17, 19.
So, the number of favourable outcomes = 4.
\( \therefore \) P(getting a prime number less than 20) = \( P(E_4) = \frac{4}{50} = \frac{2}{25} \)

Question. A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is \( \frac{3}{10} \) and that of a black ball is \( \frac{2}{5} \), then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.  
Answer: P (White ball) = \( \frac{3}{10} \)
P(Black ball) = \( \frac{2}{5} \)
P(E) = 1 - P( not E)
Probability of not getting white and black ball is equals to probability of getting red balls.
P(Red ball) = \( 1 - (\frac{3}{10} + \frac{2}{5}) = \frac{3}{10} \)
\( \frac{2}{5} \times \) Total no of balls = 20 (red balls)
Hence, Total numbers of balls = \( \frac{20 \times 5}{2} = 50 \)

Question. The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is \( \frac{1}{4} \). The probability of selecting a blue ball at random from the same jar is \( \frac{1}{3} \). If the jar contains 10 orange balls, find the total number of ball in the jar. 
Answer: P(red ball) = \( \frac{1}{4} \), P(blue ball) = \( \frac{1}{3} \)
\( \Rightarrow \) P(orange ball) = \( 1 - \frac{1}{4} - \frac{1}{3} = \frac{5}{12} \)
Suppose total no. of balls = x
Then \( \frac{10}{x} = \frac{5}{12} \)
Hence \( x = 24 \)

Question. All red face cards are removed from a pack of playing cards. The remaining cards are well-shuffled and then a card is drawn at random from them. Find the probability that the drawn card is  
i. a red card,
ii. a face card,
iii. a card of clubs.
Answer: There are 6 red face cards. These are removed.
Thus, remaining number of card = 52 – 6 = 46.
(i) Number of red cards now = 26 – 6 = 20.
Therefore, P(getting a red card) = \( \frac{\text{Number of favorable outcomes}}{\text{Number of all possible outcomes}} = \frac{20}{46} = \frac{10}{23} \).
Thus, the probability that the drawn card is a red card is \( \frac{10}{23} \).
(ii) Number of face cards now = 12 – 6 = 6.
Therefore, P(getting a face card) = \( \frac{\text{Number of favorable outcomes}}{\text{Number of all possible outcomes}} = \frac{6}{46} = \frac{3}{23} \).
Thus, the probability that the drawn card is a face card is \( \frac{3}{23} \).
(iii) The number of card of clubs = 12.
Therefore, P(getting a card of clubs) = \( \frac{\text{Number of favorable outcomes}}{\text{Number of all possible outcomes}} = \frac{12}{46} = \frac{6}{23} \).
Thus, the probability that the drawn card is a card of clubs is \( \frac{6}{23} \).

Question. A box contains cards bearing numbers 6 to 70. If one cards is drawn at random from the box, find the probability that it bears  
i. a one digit number.
ii. a number divisible by 5,
iii. an odd number less than 30,
Answer: (i) Let E be the event of getting a one digit number.
Number of possible outcomes = 70 - 6 + 1 = 65
The outcomes favourable to E are 6,7,8 and 9
\( \therefore \) Number of favourable outcomes = 4
P(E) = P(Getting a one digit number) = \( \frac{4}{65} \)
(ii) Let F be the event of getting a number divisible by 5.
Number of possible outcomes = 65
The outcomes favourable to the event F are 10,15,20,...., 65,70.
\( \therefore \) Number of outcomes favourable to F = 13
P(F) = P(Getting a number divisible by 5) = \( \frac{13}{65} = \frac{1}{5} \)
(iii) Let G be the event of getting an odd number less than 30.
Number of possible outcomes = 65
The outcomes favourable to the event G are 7,9,11,13,....., 29.
\( \therefore \) Number of favourable outcome = 12
P(G) = P(Getting an odd number less than 30) = \( \frac{12}{65} \)

Question. A box contains 90 discs which are numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears  
i. a two digit number,
ii. number divisible by 5.
Answer: No. of all possible outcomes = 90
(i) Discs with two digit number are 10 to 90
No. of discs with two digits numbers = 90 - 10 + 1 = 81
\( \therefore \) No of favourable outcomes = 81
P(a disc with two digit number) = \( \frac{\text{No. of favourable outcomes}}{\text{No. of all possible outcomes}} = \frac{81}{90} = \frac{9}{10} \)
(ii) The numbers having 0 or 5 at its once place are divisible by 5 = 5,10,15.....90
Total no. of favourable outcomes = 18
P(a disc with a number divisible by 5) = \( \frac{18}{90} = \frac{1}{5} \)

Question. A card is drawn at random from a well shuffled deck of 52 cards. The probability that it will be a spade or a king is  
(a) \( \frac{4}{13} \)
(b) \( \frac{6}{13} \)
(c) \( \frac{8}{13} \)
(d) \( \frac{10}{13} \)
Answer: (a) \( \frac{4}{13} \)

Question. The probability of an impossible event is 
(a) 0.01
(b) 100
(c) zero
(d) 1
Answer: (c) zero

Question. The probability that a leap year will have 53 Sundays or 53 Mondays is  
(a) \( \frac{4}{7} \)
(b) \( \frac{2}{7} \)
(c) \( \frac{1}{7} \)
(d) \( \frac{3}{7} \)
Answer: (d) \( \frac{3}{7} \)

Question. A box contains 3 blue balls, 2 white balls and 4 red balls. If a ball is drawn at random from the box, the probability of getting a white ball is 
(a) \( \frac{2}{9} \)
(b) \( \frac{4}{9} \)
(c) 1
(d) \( \frac{3}{9} \)
Answer: (a) \( \frac{2}{9} \)

Question. A card is drawn at random from a pack of 52 cards. Find the probability that the card is drawn is (i) a black king (ii) a jack, a queen or a king (iii) neither an ace nor a king. 
(a) (i) \( \frac{11}{13} \)(ii) \( \frac{1}{26} \)(iii) \( \frac{3}{13} \)
(b) (i) \( \frac{3}{13} \)(ii) \( \frac{1}{26} \)(iii) \( \frac{11}{13} \)
(c) (i) \( \frac{1}{26} \)(ii) \( \frac{3}{13} \)(iii) \( \frac{11}{13} \)
(d) (i) \( \frac{11}{13} \)(ii) \( \frac{3}{13} \)(iii) \( \frac{1}{26} \)
Answer: (c) (i) \( \frac{1}{26} \)(ii) \( \frac{3}{13} \)(iii) \( \frac{11}{13} \)

Question. A bag contains cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that number is divisible by both 2 and 3.  
Answer: The numbers divisible by 2 and 3 both = 6, 12, 18, 24.
Number of favourable outcomes = 4
Total outcomes = 25
Probablity of event happen P(E) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)
\( \therefore \) P(number divisible by 2 and 3) = \( \frac{4}{25} \)

Question. What is the probability that an ordinary year has 53 Sundays? 
An ordinary year has 365 days which means 52 complete weeks and one day.
If 52 weeks ends in Mon, then next day will be = Tue
If 52 weeks ends in Tue, then next day will be = Wed
If 52 weeks ends in Wed, then next day will be = Thu
If 52 weeks ends in Fri, then next day will be = Sat
If 52 weeks ends in Sat, then next day will be = Sun
If 52 weeks ends in Sun, then next day will be = Mon
Therefore,total number of outcomes = 7
Also, we have, number of cases favourable to the event = 1
Therefore, probability that an ordinary year has 53 Sundays = \( \frac{\text{number of cases favourable to the event}}{\text{total number of outcomes}} = \frac{1}{7} \).

Question. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag. 
Answer: Let the number of blue balls in the bag be x.
Then, total number of balls = (5 + x).
Given, P(a blue ball) = 3 \( \times \) P(a red ball)
\( \therefore \frac{x}{(5+x)} = 3 \times \frac{5}{(5+x)} \Rightarrow x = 15 \)

Question. A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the drawn card is neither a king nor a queen. 
Answer: Total number of cards = 52.
Total number of kings and queens = 4 + 4 = 8.
Remaining number of cards = 52 - 8 = 44.
P(getting a card which is neither a king nor a queen) = \( \frac{44}{52} = \frac{11}{13} \).

Question. Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.  
Answer: We know that, When a pair of dice is thrown, the total number of possible outcomes are 36.
Favourable outcomes of the sum greater than 10 = {(5,6),(6,5),(6,6)}
Therefore, number of cases favourable to the event = 3
Hence, Probability of getting the sum greater than 10 = \( \frac{3}{36} = \frac{1}{12} \).

Question. Two dice, one blue and one grey, are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is 8? 
Answer: Total number of outcomes = 36
Favourable outcomes are (2,6), (3,5), (4,4), (5,3), (6,2) = 5
\( \text{Probability} = \frac{\text{Number of favorable outcome}}{\text{Total number of outcome}} \)
\( \therefore \) Required probability = \( \frac{5}{36} \)

Question. A Box contains cards numbered 3,5,7,9,..,35,37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number.  
Answer: Given numbers 3, 5, 7, 9 ……, 35, 37 form an AP with a = 3 and d = 2.
Let \( T_n = 37 \). Then,
\( 3 + (n - 1)2 = 37 \)
\( \Rightarrow 3 + 2n - 2 = 37 \)
\( \Rightarrow 2n = 36 \)
\( \Rightarrow n = 18 \)
Thus, total number of outcomes = 18.
Let E be the event of getting a prime number.
Out of these numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37.
The number of favorable outcomes = 11.
Therefore, P(getting a prime number) = P(E) = \( \frac{\text{Number of outcomes favorable to E}}{\text{Number of all possible outcomes}} = \frac{11}{18} \)

Question. In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil on a card and puts them into a basket and mixes thoroughly. A child is asked to pick one card from the basket. What is the probability that the name written on the card is: 
i. the name of a girl
ii. the name of a boy?
Answer: According to the question:
Number of girls = 18
Number of boys = 16
Total pupil = 18 + 16 = 34
i. Probability of a girl monitor = \( \frac{18}{34} = \frac{9}{17} \)
ii. Probability of a boy monitor = \( \frac{16}{34} = \frac{8}{17} \)

Question. From a pack of 52 playing cards, Jacks, Queens, and Kings of red color are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is :(i) a black king, (ii) a card of red color, (iii) a card of black color. 
Answer: Total cards = 52
Cards removed = 6
Cards left = 52 - 6 = 46
So total possibilities n = 46
i. Total black kings = 2
Probability of drawing black king = \( \frac{m}{n} = \frac{2}{46} = \frac{1}{23} \)
ii. Total red cards = 26 - 6 = 20 So m = 20
iii. Probability of drawing red colour card = \( \frac{m}{n} = \frac{20}{46} = \frac{10}{23} \)
iv. Total card of black colour = 26 so m = 26
Probability of drawing black colour card = \( \frac{m}{n} = \frac{26}{46} = \frac{13}{23} \)

Question. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure), and these are equally likely out-comes. What is the probability that it will point at  
i. 8?
ii. an odd number?
iii. a number greater than 2?
iv. a number less than 9?
Answer: Total numbers = 8
\( \therefore \) Number of all possible outcomes = 8
i. Number of outcomes favourable to the event that the arrow will point at 8 = 1
\( \therefore \) Probability that the arrow will point at 8
Probability of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{8} \)
ii. Number of outcomes favourable to the event that the arrow will point at an odd number (1, 3, 5, 7) = 4
\( \therefore \) Probability that the arrow will point at an odd number
Probability of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} = \frac{4}{8} = \frac{1}{2} \)
iii. Number of outcomes favourable to the event that the arrow will point at a number greater than 2 (3, 4, 5, 6, 7, 8) = 6
\( \therefore \) Probability that the arrow will point at a number greater than 2
Probability of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} = \frac{6}{8} = \frac{3}{4} \)
iv. Number of outcomes favourable to the event that the arrow will point at a number less than 9 (1, 2, 3, 4, 5, 6, 7, 8) = 8
\( \therefore \) Probability that the arrow will point at a number less than 9 = Probability of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} = \frac{8}{8} = 1 \)

Question. Cards numbered 1 to 30 are put in a bag. A card is drawn at random. Find the probability that the drawn card is  
i. prime number > 7
ii. not a perfect square
Answer: No. of possible outcomes = 30
i. P(prime no. > 7) = 11,13,17,19,23,29 so m = 6
P(\( E_1 \)) = \( \frac{m}{n} = \frac{6}{30} = \frac{1}{5} \)
No. of perfect square are 1,4,9,16,25 = 5
ii. No. of non perfect square = 30 - 5 = 25 so m = 25
iii. P(not a perfect square) = \( \frac{m}{n} = \frac{25}{30} = \frac{5}{6} \)

Question. Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another. What is the probability that both will visit the shop on: 
i. the same day?
ii. different days?
iii. consecutive days?
Answer: Total number of days to visit the shop = 6
Two customers can visit the shop on two days in 6 \( \times \) 6 = 36 ways
So total number of outcomes = 36
i. Two customers can visit the shop on same day of the week in 6 ways i.e.
(M, M), (T, T), (W, W), (Th, Th), (F, F), (S, S)
Favourable number of ways = 6
\( \therefore \) P(both will reach on same day) = \( \frac{6}{36} = \frac{1}{6} \)
ii. We know, Probability of occurrence of an event + Probability of non occurrence of event = 1
i.e. P(E) + \( P(\overline{E}) \) = 1
P(both will reach on same day) = \( \frac{1}{6} \)
\( \Rightarrow \frac{1}{6} + P(\overline{E}) = 1 \)
\( \Rightarrow P(\overline{E}) = 1 - \frac{1}{6} \)
\( \Rightarrow P(\overline{E}) = \frac{5}{6} \)
Hence, P(both will reach on different day) = \( \frac{5}{6} \)
iii. Two customers can visit the shop on consecutive days in 5 ways i.e.
(M, T), (T, W), (W, Th), (Th, F), (F, S)
Favourable number of ways = 5
P(both will reach on consecutive days) = \( \frac{5}{36} \).

Question. Find the mean marks of the students for the following distribution 
[Marks: 0 and above, No. of Students: 80; Marks: 10 and above, No. of Students: 77; Marks: 20 and above, No. of Students: 72; Marks: 30 and above, No. of Students: 65; Marks: 40 and above, No. of Students: 55; Marks: 50 and above, No. of Students: 43; Marks: 60 and above, No. of Students: 28; Marks: 70 and above, No. of Students: 16; Marks: 80 and above, No. of Students: 10; Marks: 90 and above, No. of Students: 8; Marks: 100 and above, No. of Students: 0]
Answer: We form following table for calculations:
[Table: Marks C.I. (0-10, 10-20, ..., 100-110), \( X_i \) (5, 15, ..., 105), \( d_i=x_i-a \) (a=55) (-50, -40, ..., 50), \( f_i \) (3, 5, 7, 10, 12, 15, 12, 6, 2, 8, 0), \( f_i d_i \) (-150, -200, -210, -200, -120, 0, 120, 120, 60, 320, 0)]
a = assumed mean = 55
\( \sum f_i d_i = -260 \)
\( \sum f_i = 80 \)
\( \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} = 55 - \frac{260}{80} = 55 - \frac{13}{4} = 55 - 3.25 \)
\( \Rightarrow \overline{x} = 51.75 \)
This method is called deviation method.
Hence, the mean marks of students = 51.75.

Question. A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is : 
i. a card of a spade or an ace.
ii. a black king.
iii. neither a jack nor a king.
iv. either a king or a queen
Answer: i. Cards of spade or an ace = 13 + 3 = 16 Hence m=16
Total no. of cards = 52 So n=52
P (spade or an ace) = \( \frac{m}{n} = \frac{16}{52} = \frac{4}{13} \)
ii. Black kings = 2 so m=2
P(a black king) = \( \frac{m}{n} = \frac{2}{52} = \frac{1}{26} \)
iii. Jack or king = 4 + 4 = 8 so for neither jack nor a king = 52-8=44 hence m=44
P(neither jack nor a king) = \( \frac{m}{n} = \frac{44}{52} = \frac{11}{13} \)
iv. King or queen = 4 + 4 = 8, So m=8
P (either a king or a queen) = \( \frac{m}{n} = \frac{8}{52} = \frac{2}{13} \)

Question. A bag contains 4 white balls, 6 red balls, 7 black balls and 3 blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is 
i. white
ii. not black
iii. neither white nor black
iv. red or white.
Answer: Number of white balls in the bag = 4
Number of red balls in the bag = 6
Number of black ball in the bag = 7
Number of blue balls in the bag = 3
\( \therefore \) Total number of balls in the bag = 4 + 6 + 7 + 3 = 20
\( \therefore \) Number of all possible outcomes = 20
Proabibilty of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} \)
i. Let E be the event that the ball drawn is white.
Then, the number of outcomes favourable to E is 4.
So, P(E) = P (white) = \( \frac{4}{20} = \frac{1}{5} \)
ii. Let E be the event that the ball drawn is no black.
Then, the number of outcomes favourable to E is 4 + 6 + 3 = 13.
So, P(E) = P(not black) = \( \frac{13}{20} \)
iii. Let E be the event that the ball drawn is neither white nor black.
Then, the number of outcomes favourable to E is 6 + 3 = 9.
So, P(E) = P (neither white nor black) = \( \frac{9}{20} \)
iv. Let E be the event that the ball drawn is red or white.
Then, the number of outcomes favourable to E is 6 + 4 = 10.
So, P(E) = P (red or white) = \( \frac{10}{20} = \frac{1}{2} \)