CBSE Class 10 Probability Sure Shot Questions Set C

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1.Two coins are tossed simultaneously. Find the probability of getting

  i). at least one head

 ii). at most one head

 iii). exactly two head

 iv). exactly one head

  v). no head

 vi). no tail

vii). at least one tail

viii). at most one tail

  ix). exactly two tails

   x). exactly one tail

2. A coin is tossed two times. Find the probability of getting at most one head.

3. A coin is tossed 3 times. List the possible outcomes. Find the probability of getting (i) all heads (ii) at least 2 heads

4. Sushma tosses a coin 3 times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons.

5. If I toss a coin 3 times and get head each time, should I expect a tail to have a higher chance in the 4th toss? Give reason in support of your answer.

6. Three coins are tossed simultaneously. What is the probability of getting

  i). exactly two heads

 ii). at least two heads

 iii). at most two heads

 iv). one head or two heads

 v). exactly one tail

 vi). at least one tail

 vii). at most one tail

 viii). at least two tails

 ix). at most two tails

 x). exactly two tails

  xi). no head

 xii). no tail

7. Four coins are tossed simultaneously. What is the probability of getting

  i). exactly one head

  ii). exactly two heads

  iii). exactly three heads

  iv). at least one head

  v). at most one head 

vi). at least three heads

Question. A dice is thrown. Find the probability of getting an even number.
(a) 2/3
(b) 1
(c) 5/6
(d) 1/2
Answer: (d)
Explanation: Total number of cases = 6 (1,2,3,4,5,6)
There are three even numbers 2,4,6
Therefore probability of getting an even number is:
\( P(\text{even}) = 3/6 \)
\( \Rightarrow P(\text{even}) = 1/2 \)

Question. Two coins are thrown at the same time. Find the probability of getting both heads.
(a) 3/4
(b) 1/4
(c) 1/2
(d) 0
Answer: (b)
Explanation: Since two coins are tossed, therefore total number of cases \( = 2^2 = 4 \)
Therefore, probability of getting heads in both coins is:
\(\therefore P(\text{head}) = 1/4 \)

Question. Two dice are thrown simultaneously. The probability of getting a sum of 9 is:
(a) 1/10
(b) 3/10
(c) 1/9
(d) 4/9
Answer: (c)
Explanation: Total cases = 36
Total cases in which sum of 9 can be obtained are: (5, 4), (4, 5), (6, 3), (3, 6)
\(\therefore P(9) = 4/36 = 1/9 \)

Question. 100 cards are numbered from 1 to 100. Find the probability of getting a prime number.
(a) 3/4
(b) 27/50
(c) 1/4
(d) 29/100
Answer: (c)
Explanation: Total prime numbers from 1 to 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
That means 25 out of 100
So probability is:
\( P(\text{prime}) = 25/100 \)
\( \Rightarrow P(\text{prime}) = 1/4 \)

Question. A bag contains 5 red balls and some blue balls If the probability of drawing a blue ball is double that of a red ball, then the number of blue balls in a bag is:
(a) 5
(b) 10
(c) 15
(d) 20
Answer: (b)
Explanation: Let the number of blue balls be \( x \)
Then total number of balls will be \( 5 + x \).
According to question,
\( x/(5 + x) = 2 \times (5/5+x) \)
\( \Rightarrow x = 10 \)

Question. A box of 600 bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. Then the probability that it is non-defective bulb is:
(a) 143/150
(b) 147/150
(c) 1/25
(d) 1/50
Answer: (b)
Explanation: \( P(\text{non-defective bulb}) = 1 – P(\text{Defective bulb}) \)
\( = 1 – (12/600) \)
\( = (600 – 12)/600 \)
\( = 588/600 \)
\( = 147/150 \)

Question. Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box randomly, then the probability that the number on card is a perfect square.
(a) 9/100
(b) 1/10
(c) 3/10
(d) 19/100
Answer: (b)
Explanation: The perfect square numbers between 2 to 101 are:
4, 9, 16, 25, 36, 49, 64, 81, 100
Total numbers from 2 to 101 \( = 100 \)
So probability of getting a card with perfect square number is:
\( P(\text{perfect square}) = 10/100 \)
\( \Rightarrow P(\text{perfect square}) = 1/10 \)

Question. What is the probability of getting 53 Mondays in a leap year?
(a) 1/7
(b) 53/366
(c) 2/7
(d) 7/366
Answer: (c)
Explanation: With 366 days, the number of weeks in a year is \( 366/7 = 52 (2/7) \)
i.e., 52 complete weeks which contains 52 Mondays,
Now 2 days of the year are remaining. These two days can be
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
i.e., there are 7 pairs, in which Monday occurs in 2 pairs,
So probability is: \( P(53 \text{ Monday}) = 2/7 \)

Question. A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a king of red suit.
(a) 1/26
(b) 3/26
(c) 7/52
(d) 1/13
Answer: (a)
Explanation: There are total 4 kings in 52 cards, 2 of red colour and 2 of black colour
Therefore, Probability of getting a king of red suit is:
\( P(\text{King of red suit}) = 2/52 \)
\( \Rightarrow P(\text{King of red suit}) = 1/26 \)

Question. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number 1,2,3……12 ,then the probability that it will point to an odd number is:
(a) 1/6
(b) 1/12
(c) 7/12
(d) 5/12
Answer: (a)
Explanation: The odd numbers in 1,2,3……..12 are: 1,3,5,7,9,11
There fore probability that an odd number will come is:
\( P(\text{odd number}) = 6/12 \)
\( \Rightarrow P(\text{odd number}) = 1/2 \)

Question. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Aryan wins if all the tosses give the same result i.e. three heads or three tails and loses otherwise. Then the probability that Aryan will lose the game.
(a) 3/4
(b) 1/2
(c) 1
(d) 1/4
Answer: (a)
Explanation: Total outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
Favourable outcomes for losing game are HHT, HTH, THH, HTT, THT, TTH
There fore probability of losing the game is:
\( P(\text{Losing the game}) = 6/8 \)
\( \Rightarrow P(\text{Losing the game}) = 3/4 \)

Question. Riya and Kajal are friends. Probability that both will have the same birthday isthe same birthday is:
(a) 364/365
(b) 31/365
(c) 1/365
(d) 1/133225
Answer: (c)
Explanation: Riya may have any one of 365 days of the year as her birthday. Similarly Kajal may have any one of 365days as her birthday.
Total number of ways in which Riya and Kajal may have their birthday are: \( 365 \times 365 \)
Then Riya and Kajal may have same birthday on any one of 365 days.
There fore number of ways in which Riya and Kajal may have same birthday are: \( = 365 / (365 \times 365) = 1/365 \)

Question. A number \( x \) is chosen at random from the numbers -2, -1, 0 , 1, 2. Then the probability that \( x^2 < 2 \) is?
(a) 1/5
(b) 2/5
(c) 3/5
(d) 4/5
Answer: (c)
Explanation: We have 5 numbers \( -2, -1, 0, 1, 2 \)
Whose squares are 4, 1, 0, 1, 4
So square of 3 numbers is less than 2
There fore Probability is: \( P(x^2 < 2) = 3/5 \)

Question. A jar contains 24 marbles. Some are red and others white. If a marble is drawn at random from the jar, the probability that it is red is 2/3, then the number of white marbles in the jar is:
(a) 10
(b) 6
(c) 8
(d) 7
Answer: (c)
Explanation: Let the number of white marbles be \( x \).
Since only two colour marbles are present, and total probability we know of all the events is equal to 1.
\( P(\text{white}) = 1 – P(\text{red}) \)
\( x/24 = 1 – (2/3) \)
\( \Rightarrow x/24 = 1/3 \)
\( \Rightarrow x = 8 \)
So there are 8 white marbles.

Question. A number is selected at random from first 50 natural numbers. Then the probability that it is a multiple of 3 and 4 is:
(a) 7/50
(b) 4/25
(c) 1/25
(d) 2/25
Answer: (d)
Explanation: The numbers that are multiple of 3 (from first 50 natural numbers) are: 3, 6, 9, 12, 15, 18…………..48
The numbers that are multiple of 4 (from first 50 natural numbers) are: 4, 8, 12, 16……………….48
The numbers that are multiples of 3 and 4 both are the multiples of \( 3 \times 4 = 12 \) as both 3 and 4 are co-prime.
So common multiples are: 12, 24, 36, 48
There fore probability is: \( P(\text{multiple of 3 and 4}) = 4/50 \Rightarrow P(\text{multiple of 3 and 4}) = 2/25 \)

Question. A survey has been done on 100 people out of which 20 use bicycles, 50 use motorbikes and 30 use cars to travel from one place to another. Find the probability of persons who use bicycles, motorbikes and cars respectively?
Answer: Sol. P(of persons who use bicycles) = \( \frac{20}{100} = 0.2 \). P(of persons who use motorbikes) = \( \frac{50}{100} = 0.5 \). P(of persons who use cars) = \( \frac{30}{100} = 0.3 \).

Question. A box contains 35 blue, 25 white and 40 red marbles. If a marble is drawn at random from the box, find the probability that the drawn marble is (i) white (ii) not blue (iii) neither white nor blue.
Answer: Sol. No. of blue marbles = 35. No. of white marbles = 25. No. of red marbles = 40. Total no. of marbles = 35 + 25 + 40 = 100.
(i) P(white) = \( \frac{25}{100} = \frac{1}{4} \).
(ii) P(not a blue) = \( \frac{25 + 40}{100} = \frac{65}{100} = \frac{13}{20} \).
(iii) P(neither white nor blue) = P(red) = \( \frac{40}{100} = \frac{2}{5} \)

Question. A box contains 70 cards numbered from 1 to 70. If one card is drawn at random from the box, find the probability that it bears (i) a perfect square number. (ii) a number divisible by 2 and 3.
Answer: Sol. (i) Perfect squares upto 70 are \( 1^2, 2^2, ..., 8^2 = 8 \). \( \therefore \) P(a perfect square) = \( \frac{8}{70} = \frac{4}{35} \).
(ii) Numbers divisible by 2 and 3 are: 6, 12, 18, ..., 66, i.e., 11 nos. \( \therefore \) P(a no. divisible by 2 and 3) = \( \frac{11}{70} \)

Question. A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (i) extremely patient (ii) extremely kind or honest.
Answer: Sol. Extremely patient = 3. Extremely honest = 6. Extremely kind = 12 – 3 – 6 = 3.
(i) P(extremely patient) = \( \frac{3}{12} = \frac{1}{4} \).
(ii) P(extremely kind or honest) = \( \frac{3 + 6}{12} = \frac{9}{12} = \frac{3}{4} \)

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 drawn card is a prime number.
Answer: Sol. Total number of cards = 18. Prime numbers are: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, i.e., 11.
\( \therefore \) P(Prime number) = \( \frac{11}{18} \)

Question. Find the probability that a leap year selected at random, will contain 53 Mondays.
Answer: Sol. In a leap year, total number of days = 366. \( \therefore \) 366 days = 52 complete weeks + 2 extra days. Thus, a leap year always has 52 Mondays and extra 2 days. Extra 2 days can be, (i) Sunday and Monday, (ii) Monday and Tuesday, (iii) Tuesday and Wednesday, (iv) Wednesday and Thursday, (v) Thursday and Friday, (vi) Friday and Saturday, (vii) Saturday and Sunday. Let E be the event that a leap year has 53 Mondays. \( \therefore \) E = {Sun and Mon, Mon and Tues}. \( \therefore \) P(E) = \( \frac{2}{7} \)

Question. A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards thoroughly. Find the pro bability that the number on the drawn card is: (i) an odd number (ii) a multiple of 5 (iii) a perfect square (iv) an even prime number.
Answer: Sol. Total number of cards = 49.
(i) Odd numbers are 1, 3, 5, ......, 49, i.e., 25. \( \therefore \) P(an odd number) = \( \frac{25}{49} \).
(ii) ‘A multiple of 5’ numbers are 5, 10, 15, ......, 45, i.e., 9. \( \therefore \) P(a multiple of 5) = \( \frac{9}{49} \).
(iii) “A perfect square” numbers are 1, 4, 9, ......, 49, i.e., 7. \( \therefore \) P(a perfect square number) = \( \frac{7}{49} = \frac{1}{7} \).
(iv) “An even prime number” is 2, i.e., only one number. \( \therefore \) P(an even prime number) = \( \frac{1}{49} \)

Question. A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is (i) divisible by 2 or 3, (ii) a prime number.
Answer: Sol. (i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 3, 9, 15 = 13. Total outcomes = 20. Possible outcomes = 13. \( \therefore \) P(divisible by 2 or 3) = \( \frac{13}{20} \).
(ii) Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 = 8. Total Outcomes = 20. Possible outcomes = 8. \( \therefore \) P(a prime number) = \( \frac{8}{20} = \frac{2}{5} \)

Question. A bag contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is: (i) divisible by 3 or 5 (ii) a perfect square number.
Answer: Sol. Total number of outcomes = 25.
(i) Possible outcomes of numbers divisible by 3 or 5 in numbers 1 to 25 are (3, 6, 9, 12, 15, 18, 21, 24, 5, 10, 20, 25) = 12. \( \therefore \) P(No. divisible by 3 or 5) = \( \frac{12}{25} \).
(ii) Possible outcomes of numbers which are a perfect square = 5, i.e., (1, 4, 9, 16, 25). \( \therefore \) P(a perfect square no.) = \( \frac{5}{25} = \frac{1}{5} \)

Question. A game of chance consists of spinning an arrow on a circular board, divided into 8 equal parts, which comes to rest pointing at one of the numbers 1, 2, 3, ...., 8, which are equally likely outcomes. What is the probability that the arrow will point at (i) an odd number (ii) a number greater than 3 (iii) a number less than 9.
Answer: Sol. Total numbers = 8.
(i) “Odd numbers” are 1, 3, 5, 7, i.e., 4. \( \therefore \) P(an odd number) = \( \frac{4}{8} = \frac{1}{2} \).
(ii) “nos. greater than 3” are 4, 5, 6, 7, 8, i.e., 5. \( \therefore \) P(a number > 3) = \( \frac{5}{8} \).
(iii) “numbers less than 9” are 1, 2, 3, ...8 i.e., 8. \( \therefore \) P(a number < 9) = \( \frac{8}{8} = 1 \)

Question. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16.
Answer: Sol. X can be any one of 1, 2, 3, and 4 i.e., 4 ways. Y can be any one of 1, 4, 9, and 16 i.e., 4 ways. Total no. of cases of XY = 4 \( \times \) 4 = 16 ways. No. of cases, where product is less than 16: (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4), (4, 1) i.e., 8 ways. \( \therefore \) P (product x & y less then 16) = \( \frac{8}{16} = \frac{1}{2} \)

Question. The probability of guessing the correct answer to a certain question is \( \frac{x}{12} \). If the probability of guessing the wrong answer is \( \frac{3}{4} \), find x. If a student copies the answer, then its probability is \( \frac{2}{6} \). If he doesn’t copy the answer, then the probability is \( \frac{y}{3} \). Find the value of y.
Answer: Sol. P (guessing) + P (guessing wrong) = 1. \( \frac{x}{12} + \frac{3}{4} = 1 \Rightarrow \frac{x}{12} = 1 - \frac{3}{4} = \frac{1}{4} \Rightarrow x = \frac{12}{4} = 3 \).
P (he copies) + P (he doesn’t copy) = 1. \( \frac{2}{6} + \frac{y}{3} = 1 \Rightarrow \frac{y}{3} = 1 - \frac{2}{6} = \frac{4}{6} \Rightarrow y = \frac{4 \times 3}{6} = 2 \). \( \therefore y = 2 \).

Question. A number x is selected at random from the numbers 1, 4, 9, 16 and another number y is selected at random from the numbers 1, 2, 3, 4. Find the probability that the value of xy is more than 16.
Answer: Sol. x can be any one of 1, 4, 9, or 16, i.e., 4 ways. y can be any one of 1, 2, 3, or 4, i.e., 4 ways. Total number of cases of xy = 4 \( \times \) 4 = 16 ways. Number of cases, where product is more than 16: (9, 2), (9, 3), (9, 4), (16, 2), (16, 3), (16, 4), i.e., 6 ways. \( \therefore \) Required probability = \( \frac{6}{16} = \frac{3}{8} \)

Question. A game consists of tossing a coin 3 times and noting its outcome each time. Hanif wins if he gets three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Answer: Sol. The possible outcomes on tossing a coin 3 times are, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8. Outcomes when Hanif wins = {HHH, TTT} = 2. \( \therefore \) P (Hanif wins) = \( \frac{2}{8} = \frac{1}{4} \). \( \therefore \) P (Hanif will lose) = 1 – \( \frac{1}{4} = \frac{3}{4} \)

Question. Two dice are rolled once. Find the probability of getting such numbers on the two dice, whose product is 12.
Answer: Sol. Two dice can be thrown in 6 \( \times \) 6 = 36 ways. ‘Product is 12’ can be obtained as (2, 6), (6, 2), (3, 4), (4, 3), i.e., in 4 ways. \( \therefore \) P(Product is 12) = \( \frac{4}{36} = \frac{1}{9} \)

Question. A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be (i) a blue card (ii) not a yellow card (iii) neither yellow nor a blue card.
Answer: Sol. No. of red cards = 100. No. of yellow cards = 200. No. of blue cards = 50. Total no. of cards = 100 + 200 + 50 = 350. (i) P(a blue card) = \( \frac{50}{350} = \frac{1}{7} \). (ii) P(not a yellow card) = \( \frac{150}{350} = \frac{3}{7} \). (iii) P(neither yellow nor a blue card) = \( \frac{100}{350} = \frac{2}{7} \)

Question. Two different dice are thrown together. Find the probability that the numbers obtained have (i) even sum, and (ii) even product.
Answer: Sol. Two dice can be thrown as 6 \( \times \) 6 = 36 ways.
(i) Even sum outcomes: (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6) i.e., 18 ways. P(even sum) = \( \frac{18}{36} = \frac{1}{2} \).
(ii) Even product outcomes: (1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) i.e., 27 ways. P(even product) = \( \frac{27}{36} = \frac{3}{4} \)

Question. A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting (i) a king of red colour (ii) a face card (iii) the queen of diamonds.
Answer: Sol. (i) Total cards in a deck = 52. Total no. of kings = 4. Total no. of red kings = 2. P(a king of red colour) = \( \frac{2}{52} = \frac{1}{26} \). (ii) Face cards = 12. P(a face card) = \( \frac{12}{52} = \frac{3}{13} \). (iii) Queen of diamonds = 1 card. P(queen of diamonds) = \( \frac{1}{52} \)

Question. All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is (i) a black face card (ii) a red card.
Answer: Sol. Total no. of cards = 52. No. of cards removed = (4 + 4 + 4) = 12. Remaining cards = 40.
(i) P(a black face card) = \( \frac{2}{40} = \frac{1}{20} \).
(ii) P(a red card) = \( \frac{20}{40} = \frac{1}{2} \)

Question. Red kings and black aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is (i) a black face card (ii) a red card.
Answer: Sol. Total nos. of cards = 52. Cards removed = 2 + 2 = 4. Remaining cards = 52 – 4 = 48.
(i) P(a black face card) = \( \frac{6}{48} = \frac{1}{8} \).
(ii) P(a red card) = \( \frac{24}{48} = \frac{1}{2} \)

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: Sol. Total number of cards = 52. Black face cards = 6. Remaining cards = 52 – 6 = 46.
(i) P(Face card) = \( \frac{6}{46} = \frac{3}{23} \).
(ii) P(Red card) = \( \frac{26}{46} = \frac{13}{23} \).
(iii) P(Black card) = \( \frac{20}{46} = \frac{10}{23} \).
(iv) P(King) = \( \frac{2}{46} = \frac{1}{23} \)

Question. Cards numbered from 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: Sol. Total number of cards = 60 – 11 + 1 = 50.
(i) Odd nos. are 11, 13, 15, 17, .... 59 = 25 no. P(an odd number) = \( \frac{25}{50} = \frac{1}{2} \).
(ii) Perfect square numbers are 16, 25, 36, 49 = 4 numbers. P(a perfect square no.) = \( \frac{4}{50} = \frac{2}{25} \).
(iii) “Divisible by 5” numbers are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10 numbers. P(divisible by 5) = \( \frac{10}{50} = \frac{1}{5} \).
(iv) Prime numbers less than 20 are 11, 13, 17, 19 = 4 numbers. P(a prime no. less than 20) = \( \frac{4}{50} = \frac{2}{25} \)

Question. Red queens and black jacks 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) a king (ii) of red colour (iii) a face card (iv) a queen.
Answer: Sol. Number of red queens = 2. Number of black jacks = 2. Remaining cards = 52 – 2 – 2 = 48.
(i) P(a king) = \( \frac{4}{48} = \frac{1}{12} \).
(ii) P(red colour) = \( \frac{24}{48} = \frac{1}{2} \).
(iii) P(a face card) = \( \frac{8}{48} = \frac{1}{6} \).
(iv) P(a queen) = \( \frac{2}{48} = \frac{1}{24} \)

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: Sol. Total number of cards = 52. Red face cards = 6. Remaining cards = 52 – 6 = 46.
(i) P(red colour) = \( \frac{26 - 6}{46} = \frac{20}{46} = \frac{10}{23} \).
(ii) P(queen) = \( \frac{4 - 2}{46} = \frac{2}{46} = \frac{1}{23} \).
(iii) P(an ace) = \( \frac{4}{46} = \frac{2}{23} \).
(iv) P(a face cards) = \( \frac{12 - 6}{46} = \frac{6}{46} = \frac{3}{23} \)

Please click the link below to download CBSE Class 10 Probability Sure Shot Questions Set C.