Question 1

Five coins are tossed at a time. Then, the probability of obtaining at least one tail is

Accepted Answer

Total events = n (s) = 25 = 32 n(E) of getting heads = 1 p(E) = 1/32 ∴ n(E) = 1 - p(E) = 1 - 1/32 = 31/32

Question 2

A and B are two events such that P(A) = 0.3 and P ( A∪B) = 0.8. If A and B are independent, then P (B) is

Accepted Answer

Let P(B) = x Given, P(A∪B) = 0.8 and P(A) = 0.3 ⇒ P(A) + P(B) - P(A∩B) = 0.8 ⇒ P(A) + P(B) - P(A) P(B) = 0.8 {∵A and B are independent} ⇒ 0.3 + x - 0.3x = 0.8 ⇒ 0.7x = 0.5 ∴ x = 5/7

Question 3

A box contains 24 marbles, some are green and other are blue. If a marble is drawn at random from the box, the porbability that it is green is 2/3. The number of blue balls in the box is

Accepted Answer

Let the number of green marble = x Then, Probability of getting a green marble = xc1 / 24c1 = 2/3 ⇒ x/24 = 2/3 ⇒ x = 16

Question 4

A bag contains 3 red, 4 white and 7 black balls. The probability of drawing a red or a black ball, is

Accepted Answer

Here, total balls are 14. ∴ Required probability = (3c1 + 7c1) / 14c1 = (3 + 7)/14 = 10/14 = 5/7

Question 5

In a ward-robe, Nitish has 3 trousers. One of them is black, second is blue and third brown. In this ward-robe, he has 4 shirts also. One of them is black and the other 3 are white. He opens his ward-robe in the dark and picks out one shirt-trouser …

Accepted Answer

Probability that trousers are not black = 2/3 Probability that shirts are not black = 3/4 Required robability = 2/3 x 3/4 = 1/2

Question 6

Murari has 9 pair of dark blue socks and 9 pair of black socks. He keeps them all in the same bag. If he picks out three socks at random What is the probability that he will get a matching pair?

Accepted Answer

Since, there are only two types of socks in the bag. so, if murari pick up 3 socks then certainly two of them are of same type. Thus this is a certain event. Hence required probability = 1

Question 7

An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it ,The probability of hitting the plane at the first, second, third and fourth shots are 0.4, 0.3, 0.2, and 0.1 respectively. What is the probability that the …

Accepted Answer

First we find the probability that the plane is not hit = (1 - 0.4) ( 1 - 0.3) (1 - 0.2) (1 - 0.1) = 0.6 x 0.7 x 0.8 x 0.9 = 0.3024 ∴ probability that the plane is hit = 1 - 0.3024 = 0.6976

Question 8

Four boys and three girl stand in queue for an interview. The probability that they stand in alternate position is

Accepted Answer

Total number of possible arrangement for 4 boy and 3 girls in a queue = 7! when they occupy alternate position then the arrangement would be like BGBGBGB. Thus the total number of possible arrangement = 4! x 3! ∴ Required probability = 4! x 3! / 7! = (4 x 3 x 2 x 3 x 2) / (7 x 6 x 5 x 4 x 3 x 2) = 1/35

Question 9

An elevator starts with 5 passengers and stops at 8 different floors if the house. Find out the probability of all the 5 passengers alighting at different floors.

Accepted Answer

Required probability = favorable number of cases/ total number of =8P5 / 8 5 = 105/512

Question 10

If one cap is picked at random, what is the probability that it is either blue or yellow?

Accepted Answer

Total number of caps = 12 n(s) = 12c1 = 12 Out of ( 2 blue + 1 yellow) caps, number of ways to pick one cap n(E) = 3c1 = 3 ∴ Required probability = p(E) = n (E) / n(s) = 3/12 = 1/4

Question 11

If three marbles are picked at random, what is the probability that either all are green or all are red?

Accepted Answer

Total number of ways of selection of 3 marbles out of 12 = n(S) = 12C3 = 220 Total number of favorable events = n(E) = 3C3 + 4C3 = 1 + 4 = 5 ∴ Required probability = 5/220 = 1/44

Question 12

All the girls are not sitting together?

Accepted Answer

Required probability = 1- probability that all the girls sit together = 1 - 1/42 = 41/42

Question 13

All the boys are not sitting together?

Accepted Answer

Required probability = 1- probability that all the boys sit together = 1- 1/42 = 41/42

Question 14

All the girls are sitting together.

Accepted Answer

We have, 5 boys and 5 girls. Since, all the girls are sitting together, we consider them as one. Now, we can arrange 5 boys and 1 girl in 6! ways and these 5 girls (whom we considered as one) can also be arranged in 5! ways. ∴ Favorable number of ways = 6!5! and total number of ways to arrange 5 boys and 5 girls = 10! ∴ Required probability 6!5!/10! = 1/42

Question 15

All the boys are sitting together.

Accepted Answer

Required probability = 6!5!/10! = 1/42

Question 16

A fair coin is tossed 100 times, The probability of getting head an odd number of times is?

Accepted Answer

n(S) = 2100 n(E) = No. of favourable ways = 100C1 + 100C3 + ... 100 C99 = 2100-1 = 299 [∵ nC1 + nC3 + nC5 + ........................... = 2n-1 ] ∴ P(E)= n(E)/n(S) = 299/2100 = 1/2 Note : The given case can be generalised as "If a unbiased coin is tossed 'n' times, then the chance that the head will present itself an odd number of times is 1/2. "

Question 17

Out of 15 students studying in a class, 7 are from Maharastra, 5 are from Karnataka and 3 are from Goa, Four students are to be selected at random. What are the chances that at least one is from Karnataka?

Accepted Answer

Total possible ways of selecting 4 students out of 15 students = 15C4 = (15 x 14 x 13 x 12) / (1 x 2 x 3 x 4) = 1365 The no. of ways of selecting 4 students in which no students belongs to karnataka = 10C4 ∴ Number of ways of selecting atleast one student from karnataka = 15C4 - 10C4 = 1155. ∴ Required probability = 1155 / 1365 = 77 / 91 = 11 / 13

Question 18

In a box carrying one dozen of oranges, one third have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one oranges out of the three oranges picked up is good?

Accepted Answer

n(S) = 12C3 = (12 x 11 x 10) / (3 x 2) = 2 x 11 x 10 = 220 No. of selecting of 3 oranges out of the total 12 orange = 4C3 = 4 ∴ n(E) = No. of desired selection of oranges = 220 - 4 = 216 ∴ P(E) = n(E) / n(S) = 216 / 220 = 54/55

Question 19

What is the probability that a leap year selected randomly will have 53 Mondays?

Accepted Answer

A leap year has 366 days = 52 weeks + 2 days. These 2 days can be (Sunday, Monday, Wednesday) .... or (Saturday, Sunday). Out of these total 7 out comes there are 2 cases favourable to the desired event i, e. (Sunday, Monday) and (Monday, Tuesday) ∴ Required probability = 2/7

Question 20

What is the probability that a leap year selected randomly will have 53 Mondays?

Accepted Answer

A leap year has 366 days = 52 weeks + 2 days. The extra days 2 days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday) .... or (Saturday, Sunday). Out of these total 7 out comes there are 2 cases favorable to the desired event i,e. (Sunday, Monday) and (Monday, Tuesday) ∴ Required probability = 2/7

Probability Questions