Question 1

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

Accepted Answer

Let S be the sample space Then n(S) = no of ways of drawing 2 balls out of (6+4) = 10 C 2 10 = 10 * 9 2 * 1 =45 Let E = event of getting both balls of same colour Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4) = 6 C 2 + 4 C 2 = 6 * 5 2 * 1 + 4 * 3 2 * 1 = 15+6 = 21 Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

Question 2

A problem is given to three students whose chances of solving it are 1/2, 1/3 and 1/4 respectively. What is the probability that the problem will be solved?

Accepted Answer

Let A, B, C be the respective events of solving the problem and A , B , C be the respective events of not solving the problem. Then A, B, C are independent event ∴ A , B , C are independent events Now, P(A) = 1/2 , P(B) = 1/3 and P(C)=1/4 P A = 1 2 , P B = 2 3 , P C = 3 4 ∴ P( none solves the problem) = P(not A) and (not B) and (not C) = P A ∩ B ∩ C = P A P B P C ∵ A , B , C a r e I n d e p e n d e n t = 1 2 × 2 3 × 3 4 = 1 4 Hence, P(the problem will be solved) = 1 - P(none solves the problem) = 1 - 1 4 = 3/4

Question 3

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

Accepted Answer

We have n(s) = 52 C 2 52 = 52*51/2*1= 1326. Let A = event of getting both black cards B = event of getting both queens A∩B = event of getting queen of black cards n(A) = 52 * 51 2 * 1 = 26 C 2 = 325, n(B)= 26 * 25 2 * 1 = 4*3/2*1= 6 and n(A∩B) = 4 C 2 = 1 P(A) = n(A)/n(S) = 325/1326; P(B) = n(B)/n(S) = 6/1326 and P(A∩B) = n(A∩B)/n(S) = 1/1326 P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Question 4

A man and his wife appear in an interview for two vacancies in the same post. The probability of husband's selection is (1/7) and the probability of wife's selection is (1/5). What is the probability that only one of them is selected ?

Accepted Answer

2/7

Question 5

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Accepted Answer

Total number of outcomes possible, n(S) = 10 + 25 = 35 Total number of prizes, n(E) = 10 P ( E ) = n ( E ) n ( S ) = 10 35 = 2 7

Question 6

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Accepted Answer

Here, S = {1, 2, 3, 4, ...., 19, 20}. Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}. P(E) = n(E)/n(S) = 9/20.

Question 7

Two dice are tossed. The probability that the total score is a prime number is:

Accepted Answer

Clearly, n(S) = (6 x 6) = 36. Let E = Event that the sum is a prime number. Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) } n(E) = 15. P(E) = n(E)/n(S) = 15/36 = 5/12.

Question 8

What is the probability of getting 53 Mondays in a leap year?

Accepted Answer

1 year = 365 days . A leap year has 366 days A year has 52 weeks. Hence there will be 52 Sundays for sure. 52 weeks = 52 x 7 = 364days 366 – 364 = 2 days In a leap year there will be 52 Sundays and 2 days will be left. These 2 days can be: 1. Sunday, Monday 2. Monday, Tuesday 3. Tuesday, Wednesday 4. Wednesday, Thursday 5. Thursday, Friday 6. Friday, Saturday 7. Saturday, Sunday Of these total 7 outcomes, the favourable outcomes are 2. Hence the probability of getting 53 days = 2/7

Question 9

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

Accepted Answer

Let S be the sample space. Then, n(S) = number of ways of drawing 3 balls out of 15 = 15 C 3 = 15 * 14 * 13 3 * 2 * 1 = 455. Let E = event of getting all the 3 red balls. n(E) = 5 C 3 = 5 * 4 2 * 1 = 10. => P(E) = n(E)/n(S) = 10/455 = 2/91.

Question 10

Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6.

Accepted Answer

Clearly, n(S) = 6 x 6 = 36 Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6. Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)} n(E) = 14. Hence, P(E) = n(E)/n(S) = 14/36 = 7/18

Question 11

A letter is takenout at random from 'ASSISTANT' and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

Accepted Answer

A S S I S T A N T → A A I N S S S T T S T A T I S T I C S → A C I I S S S T T T Here N and C are not common and same letters can be A, I, S, T. Therefore Probability of choosing A = 2 C 1 9 C 1 × 1 C 1 10 C 1 = 1/45 Probability of choosing I = 1 9 C 1 × 2 C 1 10 C 1 = 1/45 Probability of choosing S = 3 C 1 9 C 1 × 3 C 1 10 C 1 = 1/10 Probability of choosing T = 2 C 1 9 C 1 × 3 C 1 10 C 1 = 1/15 Hence, Required probability = 1 45 + 1 45 + 1 10 + 1 15 = 19 90

Question 12

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

Accepted Answer

21/46

Question 13

A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

Accepted Answer

n s = C 2 30 Let A be the event of getting two oranges and B be the event of getting two non-defective fruits. and A ∩ B be the event of getting two non-defective oranges ∴ P A = C 2 20 C 2 30 , P B = C 2 22 C 2 30 a n d P A ∩ B = C 2 15 C 2 30 ∴ P A ∪ B = P A + P B - P A ∩ B = C 2 20 C 2 30 + C 2 22 C 2 30 - C 2 15 C 2 30 = 316 435

Question 14

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

Accepted Answer

Let S be the sample space. Then, n(S) = 52 C 2 =(52 x 51)/(2 x 1) = 1326. Let E = event of getting 1 spade and 1 heart. n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13 C 1 * 13 C 1 = 169. P(E) = n(E)/n(S) = 169/1326 = 13/102.

Question 15

A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

Accepted Answer

Let number of balls = (6 + 8) = 14. Number of white balls = 8. P (drawing a white ball) = 8 /14 = 4/7.

Question 16

Three unbiased coins are tossed.What is the probability of getting at least 2 heads?

Accepted Answer

Here S= {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}. Let E = event of getting at least two heads = {THH, HTH, HHT, HHH}. P(E) = n(E) / n(S) = 4/8= 1/2

Question 17

One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

Accepted Answer

Clearly, there are 52 cards, out of which there are 12 face cards. P (getting a face card) = 12/52=3/13.

Question 18

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

Accepted Answer

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize) = 1 - 16 C 1 × 14 C 1 × 12 C 1 × 10 C 1 16 C 4 = 15 39

Question 19

If two letters are taken at random from the word HOME, what is the probability that none of the letters would be vowels?

Accepted Answer

P(first letter is not vowel) = 2 4 P(second letter is not vowel) = 1 3 So, probability that none of letters would be vowels is = 2 4 × 1 3 = 1 6

Question 20

In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he. has offered English or Hindi ?

Accepted Answer

2/5

Probability Questions