CBSE Class 10 Maths Probability Worksheet Set 03

Question. Cards marked with numbers 1, 2, 3, …………, 25 are placed in a box and mixed thoroughly and one card is drawn at random from the box. The probability that the number on the card is a multiple of 3 and 5 is  
(a) \( \frac{12}{25} \)
(b) \( \frac{4}{25} \)
(c) \( \frac{1}{25} \)
(d) \( \frac{8}{25} \)
Answer: (c) \( \frac{1}{25} \)

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, 9, 10 and these values are equally likely outcomes. The probability that it will point at a number greater than 5 is  
(b) \( \frac{1}{4} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{1}{3} \)
Answer: (a) \( \frac{1}{2} \)

Question. One card is drawn from a well shuffled pack of 52 cards. The probability of getting an ace is  
(a) \( \frac{1}{52} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{4}{13} \)
(d) \( \frac{2}{13} \)
Answer: (b) \( \frac{1}{13} \)

Question. The king, queen and jack of clubs are removed from a deck of 52 cards and the remaining cards are shuffled. A card is drawn from the remaining cards. The probability of getting a king is  
(a) \( \frac{4}{52} \)
(b) \( \frac{3}{52} \)
(c) \( \frac{3}{49} \)
(d) \( \frac{4}{49} \)
Answer: (c) \( \frac{3}{49} \)

Question. Two dice are thrown simultaneously. The probability that the sum of the numbers appearing on the dice is 1 is 
(a) 3
(b) 0
(c) 2
(d) 1
Answer: (b) 0

Question. Tickets numbered from 1 to 20 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7?  
Answer: Total number of tickets = 20
{1,2,3....20}
Favourable outcomes (tickets with number as a multiple of 3 or 7) = {3,6,9,12,15,18,7,14}
Therefore,number of favourable cases to the event = 8
Required probability = \( \frac{8}{20} = \frac{2}{5} \).

Question. If three different coins are tossed together, then find the probability of getting two heads. 
Answer: All possible outcomes are : (HHH), (THH), (HTH), (HHT), (TTT), (TTH), (THT), (HTT).
Total number of outcomes = 8
No. of favourable outcomes = 3
\( \text{probability} = \frac{\text{Number of favorable outcome}}{\text{Total number of outcome}} \)
P(getting two heads) = \( \frac{3}{8} \)

Question. If P(E) = 0.20, then what is the probability of 'not E'? (1)
Answer: P(E) = 0.20
\( \therefore P(\text{not } E) = 1 - P(E) \)
= 1 - 0.20 = 0.80

Question. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?  
Answer: A = getting a rotten apple
n(S) = 900 [Total apples]
P(A) = 0.18
Let, n(A) be number of rotten apples.
Then, \( P(A) = \frac{n(A)}{n(S)} = \frac{n(A)}{900} \)
\( 0.18 \times 900 = n(A) \)
\( \therefore n(A) = 162 \)
So, there are 162 rotten apples in the heap.

Question. In tossing a die, what is the probability of getting an odd number or number less than 4?  
Answer: Odd numbers = 1,3,5,
Numbers less than 4 = 1,2,3
\( \therefore \) No. of favorable outcomes = 4
Probablity of event happen P(E) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \)
\( \therefore P(\text{an odd no. or a no. < 4}) = \frac{4}{6} = \frac{2}{3} \)

Question. A box contains 90 discs which are numbered from 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. a perfect square number
iii. a number divisible by 5.
Answer: Total number of discs = 90
\( \therefore \) Number of all possible outcomes = 90
Proabibilty of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} \)
i. Let E be the event that the disc bears a two-digit number.
Then, the number of outcomes favourable to E is 90 - 9 = 81 as from 1 to 9, the numbers are one-digited and their number is 9.
\( P(E) = P(\text{a two – digit number}) = \frac{81}{90} = \frac{9}{10} \)
ii. Let E be the event that the disc bears a perfect square number.
Then, the number of outcomes favourable to E (1, 4, 9, 16, 25, 36, 49, 64, 81) is 9.
\( \therefore P(E) = P(\text{a perfect square number}) = \frac{9}{90} = \frac{1}{10} \)
iii. Let E be the event that the disc bears a number divisible by 5.
Then, the number of outcomes favourable to E (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90) is 18.
\( \therefore P(E) = P(\text{a number divisible by 5}) = \frac{18}{90} = \frac{1}{5} \)

Question. A letter is chosen at random from the letters of the word ASSASSINATION. Find the probability that the letter chosen is an  
i. vowel
ii. consonant
Answer: There are 13 letters in the word 'ASSASSINATION' out of which one letter can be chosen in 13 ways.
\( \therefore \) Total number of elementary events = 13
i. There are 6 vowels in the word 'ASSASSINATION'. So, there are 6 ways of selecting a vowel.
\( \therefore \) Probability of selecting a vowel = \( \frac{6}{13} \)
ii. We have,
Probability of selecting a consonant
= 1 - Probability of selecting a vowel \( = 1 - \frac{6}{13} = \frac{7}{13} \)

Question. The integers from 1 to 30 inclusive are written on cards (one number on one card). These card once put in a box and well mixed. Joseph picked up one card. What is the probability that his card has (i) number 7 (ii) an even number (iii) a prime number  
Answer: Total no. of possible outcomes = 30
P(E) = Proabibilty of the event = \( \frac{\text{Number of favourble outcomes}}{\text{Total number of possible outcomes}} \)
i. P (the no.7) = \( \frac{1}{30} \)
ii. Even no. are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
Favourable outcomes = 15
Required probability, P(Even number) = \( \frac{15}{30} = \frac{1}{2} \)
iii. Prime numbers from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
No. of favourable outcomes = 10
Required probability, P(prime numbers) = \( \frac{10}{30} = \frac{1}{3} \)

Question. A box contains cards, number 1 to 90. A card is drawn at random from the box. Find the probability that the selected card bears a : 
i. Two digit number.
ii. Perfect square number.
Answer: No. of all possible outcomes = 90
i. Total no of cards having 2 digit number = 90 - 9 = 81
P(card with two digit number) = \( \frac{81}{90} = \frac{9}{10} \)
ii. Perfect square numbers between 1 to 90 are 1,4,9,16,25,36,49,64,81
P(card with perfect square number) = \( \frac{9}{90} = \frac{1}{10} \)

Question. Figure show the top view of an open square box that is divided into 6 compartments with walls of equal height. Each of the rectangles D, E, F has twice the area of each of the squares A, B and C. When a marble is dropped into the box at random, it falls into one of the compartments. What is the probability that it will fall into compartment F? 
Answer: An open square box that is divided into 6 compartments with walls of equal height. Each of the rectangles D, E, F has twice the area of each of the squares A, B and C. Therefore
Let x square units be the area of the upper face of each of the compartments A, B and C. Then, area of the upper face of each compartment D, E and F is 2x sq. units.
Area of the square box = (x + x + x + 2x + 2x + 2x) sq. units = 9x sq. units.
P (Marble falls in compartment F) = \( \frac{\text{Area of compartment F}}{\text{Area of square box}} = \frac{2x}{9x} = \frac{2}{9} \)

Question. Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25? 
Answer: The person having higher probability of getting the number 25 has the better chance.
When a pair of dice is thrown, there are 36 elementary events which are as follows:
(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)
Therefore, the product of numbers on two dice can take values 1, 2, 3, ..., 36.
We observe that the product of two numbers on two dice will be 25 if both the dice show number 5. Therefore, there is only one elementary event, viz., (5, 5), which is favourable for getting 25.
\( p_1 = \) Probability that Peter throws 25 = \( \frac{1}{36} \)
Rina throws a die on which she can get any one of the six numbers 1, 2, 3, 4, 5, 6 as an outcome. If she gets number 5 on the upper face of the die thrown, then the square of the number is 25.
\( p_2 = \) Probability that the square of number obtained is 25 = \( \frac{1}{6} \)
Therefore, \( p_2 > p_1 \). Therefore, Rina has better chance to get the number 25.

Question. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting  
a. Non face card,
b. Black king or a Red queen,
c. Spade card.
Answer: Total number of cards n = 52
Number of face cards = 12
So no. of non face cards m = 52 - 12 = 40
P(non-face cards) = \( \frac{m}{n} = \frac{40}{52} = \frac{10}{13} \)
Number of black kings = 2
Number of red queens = 2
So m = 2 + 2 = 4
P(a black King or a red queen) = \( \frac{m}{n} = \frac{4}{52} = \frac{1}{13} \)
Number of spade cards m = 13
So P(Spade cards) = \( \frac{m}{n} = \frac{13}{52} = \frac{1}{4} \)

Question. All the black face cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a  
i. face card,
ii. red card,
iii. black card,
iv. king
Answer: Out of 52 playing cards; 2 black jacks, 2 black queens and 2 black kings have been removed.
Total number of remaining cards = (52 - 6) = 46.
i. Now, there are 6 face cards in the remaining cards.
\( \therefore \) P(getting a face card) = \( \frac{6}{46} = \frac{3}{23} \)
ii. There are 26 red cards.
\( \therefore \) P(getting a red card) = \( \frac{26}{46} = \frac{13}{23} \)
iii. Out of 46 cards, number of black cards = 26 - 6 = 20.
\( \therefore \) P(getting a black card) = \( \frac{20}{46} = \frac{10}{23} \)
iv. Now, these 46 cards have 2 kings.
P(getting a king) = \( \frac{2}{46} = \frac{1}{23} \)

Question. All the red face cards are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is:  
i. of red colour
ii. a queen
iii. an ace
iv. a face card.
Answer: No. of cards removed = 3 face cards of heart + 3 face cards of diamond = 6
Remaining cards = 52 - 6 = 46
So total No. of events are n = 46
i. No. of red card left = 13 - 6 = 7 so m = 7
so P(E) = \( \frac{7}{46} \)
ii. No. of queen left = 4 - 2 queens of heart and diamond = 2
So m = 2
P (E) = \( \frac{2}{46} = \frac{1}{23} \)
iii. Total No. of aces = 4 so m = 4
P(E) = \( \frac{4}{46} = \frac{2}{23} \)
iv. No. of face cards left = 12 - total face cards removed = 12 - 6 = 6
So m = 6
Hence P(E) = \( \frac{6}{46} = \frac{3}{23} \)

Question. A bag contains 15 balls of which x are blue and the remaining are red. If the number of red balls are increased by 5, the probability of drawing the red balls doubles. Find :  
i. P(red ball)
ii. P(blue ball)
iii. P(blue ball if of 5 extra red balls are actually added)
Answer: initially \( P(\text{red ball}) = \frac{15-x}{15} \)
and after adding 5 red balls \( P(\text{red ball}) = \frac{20-x}{20} \) So as per the question
\( \frac{20-x}{20} = 2 \times \left( \frac{15-x}{15} \right) \)
\( \frac{20-x}{4} = 2 \times \left( \frac{15-x}{3} \right) \)
60 - 3x = 120 - 8x
5x = 60
x = 12
So Blue balls = 12 and red balls = 15 - 12 = 3
i. P(red ball) = \( \frac{3}{15} = \frac{1}{5} \)
ii. P(blue ball) = \( \frac{12}{15} = \frac{4}{5} \)
iii. P(blue ball if 5 red balls are added) = \( \frac{12}{20} = \frac{3}{5} \)

Question. A ticket is drawn from a bag containing 100 tickets numbered from 1 to 100. The probability of getting a ticket with a number divisible by 10 is
(a) \( \frac{3}{10} \)
(b) \( \frac{1}{10} \)
(c) \( \frac{4}{10} \)
(d) \( \frac{1}{5} \)
Answer: (b) \( \frac{1}{10} \)
Explanation: Number of possible outcomes = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100} = 10. Number of Total outcomes = 100. Required Probability = \( \frac{10}{100} = \frac{1}{10} \).

Question. 3 rotten eggs are mixed with 12 good ones. One egg is chosen at random. The probability of choosing a rotten egg is
(a) \( \frac{1}{15} \)
(b) \( \frac{4}{5} \)
(c) \( \frac{1}{5} \)
(d) \( \frac{2}{5} \)
Answer: (c) \( \frac{1}{5} \)
Explanation: Number of possible outcomes = 3. Number of Total outcomes = 15. Required Probability = \( \frac{3}{15} = \frac{1}{5} \).

Question. A letter of English alphabets is chosen at random. The probability that the letter chosen is a vowel is
(a) \( \frac{2}{26} \)
(b) \( \frac{4}{26} \)
(c) \( \frac{1}{26} \)
(d) \( \frac{5}{26} \)
Answer: (d) \( \frac{5}{26} \)
Explanation: We know that "A, E, I, O, U" are vowels. Number of vowels = 5. Number of possible outcomes = 5. Number of total outcomes = 26. Required Probability = \( \frac{5}{26} \).

Question. If S is the sample space of a random experiment, then P(S) =
(a) \( \frac{1}{4} \)
(b) \( \frac{1}{8} \)
(c) 1
(d) 0
Answer: (c) 1
Explanation: If S is the sample space of a random experiment, then P(S) = 1.

Question. A bag contains 6 red, 8 white, 4 green and 7 black balls. One ball is drawn at random. The probability that the ball drawn is neither green nor white is
(a) \( \frac{8}{25} \)
(b) \( \frac{12}{25} \)
(c) \( \frac{13}{25} \)
(d) \( \frac{4}{25} \)
Answer: (c) \( \frac{13}{25} \)
Explanation: Total number of balls = 25. Number of Green and White balls = 4 + 8 = 12. Number of balls neither green nor white = 25 - 12 = 13. Number of possible outcomes = 13. Number of total outcomes = 25. Required Probability = \( \frac{13}{25} \).

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 and these are equally likely outcomes. Find the probability that the arrow will point at any factor of 8?
Answer: Total number of points = 8. Total number of possible outcomes = 8. Possible outcomes = \( (1 \times 8), (2 \times 4), (8 \times 1), (4 \times 2) \). No. of favourable outcomes = 4. Probability of event happen P(E) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \). \( \therefore P(\text{Factor of 8}) = \frac{4}{8} = \frac{1}{2} \).

Question. An unbiased die is thrown. What is the probability of getting a number between 3 and 6?
Answer: The event "Getting a number between 3 and 6" occurs if we obtain either 4 or 5 as an outcome. Favourable number of outcomes = 2. Hence, required probability = \( \frac{2}{6} = \frac{1}{3} \).

Question. A bag contains 18 balls out of which x balls are red. If one ball is drawn at random from the bag, what is the probability that it is not red?
Answer: The total number of balls = 18. Number of red balls = x. Number of balls which are not red = 18 – x. Therefore, P(getting a ball which is not red) = \( \frac{\text{Number of favorable outcomes}}{\text{Number of all possible outcomes}} = \frac{18 - x}{18} \). Thus, the probability of drawing a ball which is not red is \( \frac{18 - x}{18} \).

Question. A die is thrown. Find the probability of getting a number lying between 2 and 6.
Answer: Favourable outcomes of getting a number lying between 2 and 6 = {3, 4, 5}. Therefore, number of favourable outcomes = 3. Hence, Probability of getting a number lying between 2 and 6 = \( \frac{\text{number of favourable outcomes}}{\text{number of total outcomes}} = \frac{3}{6} = \frac{1}{2} \).

Question. 20 tickets, on which numbers 1 to 20 are written, are mixed thoroughly and then a ticket is drawn at random out of them. Find the probability that the number on the drawn ticket is a multiple of 3 or 7.
Answer: Total number of cases = 20. \( n(s) = 20 \). A = favourable cases = {3, 6, 7, 9, 12, 14, 15, 18}. \( n(A) = 8 \). Probability of event happen P(E) = \( \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} \). Required probability = \( P(A) = \frac{n(A)}{n(S)} = \frac{8}{20} = \frac{2}{5} \).

Question. It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb?
Answer: Out of 600 electric bulbs one bulb can be chosen in 600 ways. Total number of elementary events = 600. There are 588 (= 600 - 12) non-defective bulbs out of which one bulb can be chosen in 588 ways. Favourable number of elementary events = 588. Hence, P (Getting a non-defective bulb) = \( \frac{588}{600} = \frac{49}{50} = 0.98 \).

Question. From a group of 3 boys and 2 girls we select two children. What is the set representing the event:
i. one girl is selected
ii. at least one girl is selected?
Answer: Let boys be \( B_1, B_2, B_3 \) (3 Boys). Let girls be \( G_1, G_2 \) (2 girls). Therefore, the set which represents, one girl is selected and at least one girl is selected are respectively as,
i. \( \{B_1G_1, B_2G_1, B_3G_1, B_1G_2, B_2G_2, B_3G_2\} \)
ii. \( \{B_1G_1, B_2G_1, B_3G_1, B_1G_2, B_2G_2, B_3G_2, G_1G_2\} \)

Question. A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card is neither a red card nor a queen.
Answer: Total number of red cards = 26 (including 2 queens). Total number of queen in pack of 52 cards is 4, out of which 2 are black queen cards and 2 are red queen cards. Total number of red cards and queen cards = 26 + 2 = 28. Number of favourable outcomes = 52 - 28 = 24. \( \therefore \) P(neither red nor queen) = \( \frac{24}{52} = \frac{6}{13} \).

Question. A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. A ball is drawn at random from the bag. Find the probability that it is
i. black,
ii. not green,
iii. red or white,
iv. neither red nor green.
Answer: Total number of balls = 4 + 5 + 2 + 4 = 15.
i. Number of black balls = 2. P(getting a black ball) = \( \frac{2}{15} \).
ii. Number of balls which are not green = 4 + 5 + 2 = 11. P(getting a ball which is not green) = \( \frac{11}{15} \).
iii. Number of balls which are red or white = 5 + 4 = 9. P(getting a ball which is red or white) = \( \frac{9}{15} = \frac{3}{5} \).
iv. Number of balls which are neither red nor green = 4 + 2 = 6. P(getting a ball which is neither red nor green) = \( \frac{6}{15} = \frac{2}{5} \).

Question. A bag contains 5 red and some blue balls,
i. if probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag.
ii. if probability of drawing a blue ball from the bag is four times that of a red ball, find the number of blue balls in the bag.
Answer: Let number of blue balls = \( x \). Total number of balls = \( 5 + x \). Probability of red ball = \( \frac{5}{5+x} \). Probability of blue ball = \( \frac{x}{5+x} \).
By given condition,
i. \( \frac{x}{5+x} = 2(\frac{5}{5+x}) \). \( x = 10 \). No. of blue balls = 10.
ii. Here, \( \frac{x}{5+x} = 4(\frac{5}{5+x}) \). \( x = 20 \). Hence, the number of blue balls = 20.

Question. In a game, the entry fee is Rs 5. The game consists of tossing a coin 3 times. If one or two heads show, Shweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. For tossing a coin three times, find the probability that she
i. loses the entry fee.
ii. gets double entry fee.
iii. just gets her entry fee back.
Answer: Possible outcomes when a coin is tossed 3 times: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT. Total number of outcomes = 8.
i. Shweta will lose the entry fee if she gets 'TTT'. P(Shweta losses the entry fee) = \( \frac{1}{8} \).
ii. Shweta will get double the entry fee if she gets HHH. P(Shweta will get double the entry fee) = \( \frac{1}{8} \).
iii. Shweta will get her entry fee back if she gets HHT, HTH, THH, TTH, THT or HTT. P(Shweta will get her entry fee) = \( \frac{6}{8} = \frac{3}{4} \).

Question. Cards marked with numbers 1, 3, 5, ..., 101 are placed in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the number on the drawn cards is
i. less than 19,
ii. a prime number less than 20.
Answer: Given numbers 1, 3, 5, . . . . . . , 101 form an AP with a = 1 and d = 2. (first term is one and common difference is two). Let \( T_n = 101 \). Then, \( 1 + (n – 1)2 = 101 \Rightarrow 1 + 2n – 2 = 101 \Rightarrow 2n = 102 \Rightarrow n = 51 \). Therefore, total number of outcomes = 51.
i. Suppose \( E_1 \) be the event of getting a number less than 19. Out of these numbers, less than 19 are 1, 3, 5, ....... , 17. Given number 1, 3, 5, ......., 17 form an AP with a = 1 and d = 2. Suppose \( T_n = 17 \). Then, \( 1 + (n – 1)2 = 17 \Rightarrow 1 + 2n – 2 = 17 \Rightarrow 2n = 18 \Rightarrow n = 9 \). Thus, number of favorable outcomes = 9. Therefore, \( P(E_1) = \frac{\text{Number of outcomes favorable to } E_1}{\text{Number of all possible outcomes}} = \frac{9}{51} = \frac{3}{17} \).
ii. Suppose \( E_2 \) be the event of getting a prime number less than 20. Out of these numbers, prime numbers less than 20 are 3, 5, 7, 11, 13, 17 and 19. Therefore, the number of favorable outcomes = 7. Therefore, \( P(E_2) = \frac{7}{51} \).

Question. From a pack of 52 playing cards jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is
i. a black queen
ii. a red card
iii. a ten
iv. a picture card (jacks, queens and kings are picture cards).
Answer: There will be 52 cards in a deck. There are four different suits: Diamonds, Clubs, Hearts, and Spades. There will be thirteen cards in each suit, they are: Ace, 2,3,4,5,6,7,8,9,10, Jack, Queen, King. From a pack of 52 cards jacks, queens, kings and aces of red colour are removed. Number of cards removed = 2 + 2 + 2 + 2 = 8. Total number of remaining cards = 52 - 8 = 44. Now, there are 2 jacks, 2 queens, 2 kings and 2 aces of black colour only.
i. Number of black queens = 2. \( \therefore \) P(getting a black queen) = \( \frac{2}{44} = \frac{1}{22} \).
ii. Remaining number of red cards = 26 - 8 = 18. \( \therefore \) P(getting a red card) = \( \frac{18}{44} = \frac{9}{22} \).
iii. Number of tens = 4. \( \therefore \) P(getting ten) = \( \frac{4}{44} = \frac{1}{11} \).
iv. We know that jacks, queens and kings are picture cards. Out of 12 picture cards, it is given that 6 have been removed. So, the remaining number of picture cards = 12 - 6 = 6. \( \therefore \) P(getting a picture card) = \( \frac{6}{44} = \frac{3}{22} \).

Question. All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is drawn at random from the remaining pack. Find the probability of getting
i. a black face card
ii. a queen
iii. a black card
iv. a spade
Answer: Here, all face cards of spades are removed from a deck of 52 playing cards. So, remaining cards in deck = 52 - 3 = 49. Total number of outcomes n = 49.
i. We know that there are 6 black face cards in a deck of cards. After removing face cards of spades only 3 face cards of club are left. So number of favorable outcomes m = 3. Required Probability = \( P(E) = \frac{m}{n} = \frac{3}{49} \).
ii. There are 4 queens in a deck. After removing a queen of spade, we are left with 3 queens. Then, number of favorable outcomes m = 3. Required Probability = \( P(E) = \frac{m}{n} = \frac{3}{49} \).
iii. There are 26 black cards in a regular deck of cards. After removing 3 face cards of spades, there are only 23 black cards. Then, number of favorable outcomes m = 23. Required Probability = \( P(E) = \frac{m}{n} = \frac{23}{49} \).
iv. There are 13 cards of spade in a deck. After removing 3 face cards of spade only 10 spades cards are left. So number of favorable outcomes m = 10. Required Probability = \( P(E) = \frac{m}{n} = \frac{10}{49} \).

Question. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses proceeding the house numbered X is equal to sum of the numbers of houses following X.
Answer: The houses are numbered consecutively from 1 to 49. 1, 2, 3....................(x - 1), x, (x+1)..........49. Sum of number of houses preceding x numbered house = Sum of number following x. Sum of number of houses preceding x numbered house = \( S_1 = \frac{x-1}{2} \times (1 + x - 1) = \frac{x(x-1)}{2} \). Sum of number following x = \( S_2 = (1 + 2 + 3 + . . . . . . . 49) - \frac{x}{2} (x + 1) = \frac{49 \times 50}{2} - \frac{x^2 + x}{2} = \frac{2450 - x^2 - x}{2} \). As \( S_1 = S_2 \), \( \frac{2450 - x^2 - x}{2} = \frac{x(x - 1)}{2} \). \( 2450 - x^2 - x = x^2 - x \). \( 2x^2 = 2450 \). \( x^2 = 1225 \). \( x = 35 \). Hence sum of numbers of houses proceeding the house numbered 35 is equal to sum of the numbers of houses following 35.