Question 1

If three marbles are picked at random, what is the probability that two are blue and one is yellow?

Accepted Answer

Ways of selection of two blue marbles = 4C2 Ways of selection of one yellow marble = 3C1 Ways of selection of three marbles = 15C3 So, required probability = (4C2 x 3C1) / (15C3) = [4!/{2! (4 - 2)!} x 3!/{1! (3 - 1)!}] / [15! / {3! (15 - 3)!}] = [(4 x 3 x 2 x 1) / (2 x 1 x 2 x 1)] x [(3 x 2 x 1) / (1 x 2 x 1)] / [ (15 x 14 x 13 x 12!) / (3 x 2 x 12!)] = (6 x 3) / (5 x 7 x 13) = 18 / 455

Question 2

If four marbles are picked at random, what is the probability that at least one is blue?

Accepted Answer

Probability that none is blue from four marbles = 15 - 4C4 / 15C4 = 11C4 / 15C4 = 22/91 So, probability that at least one is blue from four marbles = 1 - 22/91 = 69/91

Question 3

If two marbles are picked at random, what is the probability that either both are green or both are yellow?

Accepted Answer

Ways of selection of two green marbles = 2C2 = 1 Ways of selection of two yellow marbles = 3C2 = 3 So, probability (both are green) = 1/15C2 ....(I) Probability (both are yellow) = 3/15C2 ....(Ii) Then, required probability = 1/15C2 + 3/15C2 = 4/ 15C2 = 4/105

Question 4

If four marbles are picked at random, what is the probability that one is green, two are blue and one is red?

Accepted Answer

Ways of selection of 4 marbles = n(S) = 15C4 Ways of selection of one green marbles = n(E1) = 2C1 Ways of selection of two blue marbles = n(E2)=4C2 Ways of selection of one red marble = n(E3) = 6C1 ∴ Required probability = (2C1 x 4C2 x 6C1) / (15C4) = 24/455

Question 5

If two caps are picked at random, what is the probability that both are blue?

Accepted Answer

Total number of caps = 2 + 4 + 5 + 1 = 12 Total number of outcomes = n(S) = 12 C2 = 66 Favorable number of outcomes = n(E) = 2C2 = 1 ∴ Required probability = 1/66

Question 6

If four caps are picked at random, what is the probability that none is green?

Accepted Answer

Total number of caps = 12 Total number of result = n(S) = 12C4 = 495 Out of 5 caps, number of ways to not pick a green cap = n(E1) = 5C0 = 1 and out of 7 caps, number of ways to pic 4 caps = n(E) = 7C4 = 35 ∴ Required probability = (1 x 35)/495 = 7/99

Question 7

If three caps are picked at random, what is the probability that two are red and one is green?

Accepted Answer

Total number of caps = 12 ∴ n(S) = 12C3 = 220 n(E1) = Out of 4 red caps, number of ways to pick 2 caps = 4C2 = 6 n(E2) = Out of 5 green caps Number of ways to pick one cap = 5C1 = 5 P(E) = n(E1) x n(E2)/n(S) = (6 x 5)/220 = 3/22

Question 8

If two caps are picked at random, what is the probability that atleast one is red?

Accepted Answer

Required probability = 1 - 8C2/ 12C2 = 1 - 28/66 = 38/66 = 19/33

Question 9

If two marbles are drawn at random, what is the probability that both are red?

Accepted Answer

Total number of possible outcomes = n(S) = 12C2 = (12 x 11)/2 = 66 Total number of favorable events = n(E) = 4C2 = (4 x 3) / (2 x 1) = 6 ∴ Required probability = 6/66 = 1/11

Question 10

If three balls are picked at random, what is the probability that atleast one is red?

Accepted Answer

Total Number of outcomes = 10C3 = 120 Number of outcomes not containing red balls = 5C3 = 10 ∴ Probability that at least one is red = 1 - 10/120 = 11/12

Question 11

If four balls are picked at random, what is the probability that two are green and two are blue?

Accepted Answer

Total number of outcomes = 10c4 = 210 Favourable number of outcomes = 3c2 x 2c2 = 3 x 1 = 3 ∴ Required probability = 3/210 = 1/70

Question 12

Two cards are drawn at random from a pack of 52 cards. What is the probability that the drawn cards are both aces?

Accepted Answer

n(S) = Number of ways of drawing 2 cards out of 52 52C2 = (52 x 51) / (2 x 1) = 1326 n(E) = Number of ways of drawing 2 cards out of 4 = 4C2 = (4 x 3)/2 = 6 ∴ P(E) = 6/1326 = 1/221

Question 13

A card is drawn at random from a pack of 52 cards. A card is drawn at random. What is the probability that it is neither a heart nor a king?

Accepted Answer

There are 13 hearts and 3 more kings ∴ P(heart or a king) = (13 + 3)/52 = 4/13 ∴ P (neither a heart nor a king) = 1 - 4/13 = 9/13

Question 14

if from each of the three boxes containing 3 white and 1 black, 2 white and 2 black and 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is?

Accepted Answer

You can have W, W, B or W, B, W or B, W, W Reqd. probability = 3/4. 2/4. 3/4 + 3/4. 2/4. 1/4 + 1/4. 2/4. 1/4 = 9/32 + 3/32 + 1/32 = 13/32

Question 15

A box contains 3 white and 2 red balls. If we draw one ball and ball and without replacing the first ball. The probability of drawing red ball in the second drawn is?

Accepted Answer

Total balls in the box = 5 Second red ball can be drawn in two ways Case I. First ball is white and second ball is red. Its probability = 3/5.2/4 = 6/20 = 3/10 Case II: First ball is red and second ball is red Its probability = 2/5.1/4 = 2/20 = 1/10 Hence, reqd. probability = 3/10 + 1/10 = 4/10 = 2/5

Question 16

Three mangoes and three apples are kept in a box. if two fruits are chosen at random. find the probability that one is a mango and the other is an apple.

Accepted Answer

Total number of ways = n(s) = 6c2 = 15 Favorable number of ways = n(E) = 3C1 x 3C1= 9 ∴ Required probability = 9/15 = 3/5

Question 17

The probability that a man will be alive for 10 more years is 1/4 and the probability that his wife will alive for 10 more years is 1/3 .The probability that none of them will be alive for 10 more years, is

Accepted Answer

Required probability = P(A) x P(B) = (1 - 1/4) x (1 - 1/3) = 3/4 x 2/3 = 1/2

Question 18

In a lottery 10000 tickets are sold and ten prizes are awarded. What is the probability of not getting a prize, if you buy one ticket?

Accepted Answer

Total lottery tickets = 10000 Total prize in the lottery = 10 ∴ Probability getting a prize = 10/10000 = 1/1000 Now, probability of not getting a prize = 1 - probability of getting a prize = 1 - (1/1000) = 999/1000

Question 19

Tow persons A and B appear in an interview for two vacancies. If the probabilities of their selections are 1/3 and 1/6 respectively, then the probability that none of them is selected, is

Accepted Answer

Required probability = P (A) x p (B) = (1- 1/4) x (1- 1/6) = (3/4) x (5/6) = 5/8

Question 20

Out of 13 applicants for a job, there are 5 women and 8 men It is desired to select 2 persons for the job, The probability that atleast one of the selected persons will be a woman, is

Accepted Answer

Total ways = 13c2 Favourable number ways of selecting men only = 8c2 ∴ Probability of selecting no woman = 8c2 / 13c2 = 14/39 ∴ Probability of selecting at least one woman = 1 - (14 / 39) = 25 / 39

Probability Questions