Question 1

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

Accepted Answer

Total number of balls = (2 + 3 + 2) = 7. Let S be the sample space. Then, n(S) = Number of ways of drawing 2 balls out of 7 = 7 C 2 = 21 Let E = Event of drawing 2 balls, none of which is blue. n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls = 5 C 2 = 10 Therefore, P(E) = n(E)/n(S) = 10/ 21.

Question 2

An unbiased die is tossed.Find the probability of getting a multiple of 3.

Accepted Answer

Here S = {1,2,3,4,5,6} Let E be the event of getting the multiple of 3 Then, E = {3,6} P(E) = n(E)/n(S) = 2/6 = 1/3

Question 3

In a bag which contains 40 balls, there are 18 red balls and some green and blue balls. If two balls are picked up from the bag without replacement, then the probability of the first ball being red and second being green is 3/26. Find the number of …

Accepted Answer

Total balls = 40 Red balls = 18 Let green balls are x Then, (18/40) × (

Question 4

There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

Accepted Answer

Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways. Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways. Required probability =24/64 = 3/8

Question 5

In a race, the odd favour of cars P,Q,R,S are 1:3, 1:4, 1:5 and 1:6 respectively. Find the probability that one of them wins the race.

Accepted Answer

P ( P ) = 1 4 , P ( Q ) = 1 5 , P ( R ) = 1 6 , P ( S ) = 1 7 All the events are mutually exclusive hence, Required probability = P(P)+P(Q)+P(R)+P(S) 1 4 + 1 5 + 1 6 + 1 7 = 319 420

Question 6

14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.

Accepted Answer

Total no of ways = (14 – 1)! = 13! Number of favorable ways = (12 – 1)! = 11! So, required probability = 11 ! × 3 ! 13 ! = 39916800 × 6 6227020800 = 24 625

Question 7

What is the probability that a leap year has 53 Saturdays and 52 Sundays ?

Accepted Answer

A leap year has 52 weeks and two days Total number of cases = 7 Number of favourable cases = 1 i.e., {Friday, Saturday} Required Probability = 1/7

Question 8

A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the ti…

Accepted Answer

As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order). The overall resultant will remain same. So final amount with the person will be (in all cases): = 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27 Hence the final result is: 64 − 27 = 37 A loss of Rs.37

Question 9

Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?

Accepted Answer

Required probability is given by P(E) = n ( E ) n ( S ) = 1 4 C 2 = 1 6

Question 10

An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.

Accepted Answer

Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then Required probability = P A ∩ B ∩ C = P A P B A P C A ∩ B Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9 When a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white ∴ P B A = 3 17 Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white. ∴ P C A ∩ B = 2 16 = 1 8 Hence the required probability = 2 9 × 3 17 × 1 8 = 1 204

Question 11

Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.

Accepted Answer

Let A be the event that X is selected and B is the event that Y is selected. P(A) = 1/7, P(B) = 2/9. Let C be the event that both are selected. P(C) = P(A) × P(B) as A and B are independent events: = (1/7) × (2/9) = 2/63

Question 12

Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King

Accepted Answer

Let X be the event that cards are in a club which is not king and other is the king of club. Let Y be the event that one is any club card and other is a non-club king. Hence, required probability: =P(A)+P(B) = 12 C 1 * 1 C 1 52 C 2 + 13 C 1 * 3 C 1 52 C 2 = 12 * 2 52 * 51 + 13 * 3 * 2 52 * 51 = 24 + 78 52 * 51 = 1 26

Question 13

A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?

Accepted Answer

Given number of balls = 3 + 5 + 7 = 15 One ball is drawn randomly = 15C1 probability that it is either pink or red = 7 C 1 + 3 C 1 15 C 1 = 7 + 3 15 = 10 15 = 2 3

Question 14

When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?

Accepted Answer

35/36

Question 15

A coin is tossed 5 times. What is the probability that head appears an odd number of times?

Accepted Answer

The possible outcomes are as follows : 5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all. Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

Question 16

A five digit number is formed with the digits 0,1,2,3 and 4 without repetition. Find the chance that the number is divisible by 5.

Accepted Answer

5 digit number = 5! = 120 Divisible by 5 then the last digit should be 0 Then the remaining position have the possibility = 4! = 24

Question 17

What is the probability of an impossible event?

Accepted Answer

The probability of an impossible event is 0. The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen. The probability of a certain event is 1.

Question 18

What is the probability of getting at least one six in a single throw of three unbiased dice?

Accepted Answer

Find the number of cases in which none of the digits show a '6'. i.e. all three dice show a number other than '6', 5×5×5=125 cases. Total possible outcomes when three dice are thrown = 216. The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'. =216−125=91 The required probability = 91/256

Question 19

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

Accepted Answer

P( only one of them will be selected) = p[(E and not F) or (F and not E)] = P E ∩ F ∪ F ∩ E = P E P F + P F P E = 1 7 × 4 5 + 1 5 × 6 7 = 2 7

Question 20

If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.

Accepted Answer

Total number of elementary events = 10 C 5 Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is 7 C 5 . So,required probability = 7 C 5 / 10 C 5 = 1/12.

Probability Questions