Question 1

An unbiased cubic die is thrown.What is the probabiltiy of getting a multiple of 3 or 4?

Accepted Answer

Total numbers in a die=6 P(mutliple of 3) = 2/6 = 1/3 P(multiple of 4) = 1/6 P(multiple of 3 or 4) = 1/3 + 1/6 = 1/2

Question 2

One card is drawn from a pack of 52 cards,each of the 52 cards being equally likely to be drawn.Find the probability that the ball drawn is an ace :

Accepted Answer

Total number of events=52 Number of aces =4 So,required probability = 4/52 =1/13

Question 3

The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?

Accepted Answer

Let A be the event of A will tell truth. B be the event of B tell truth P ( A ) = 2 3 , P ( A c ) = 1 - P ( A ) = 1 3 P ( B ) = 4 5 , P ( B c ) = 1 - P ( B ) = 1 5 When both agree then they say true or they say false together, that is A ∩ B o r A c ∩ B c Also these events will be mutually exclusive : P ( A ∩ B ) + P ( A c ∩ B c ) = P ( A ) P ( B ) P ( C ) = 2 3 * 4 5 + 1 3 * 1 5 = 3 5

Question 4

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

Multiples of 3 below 20 are 3, 6, 9, 12, 15, 18 Multiples of 5 below 20 are 5, 10, 15, 20 Required number of possibilities = 10 Total number of possibilities = 20 Required probability = 10/20 = 1/2.

Question 5

Three fair coins are tossed simultaneously. Find the probability of getting more heads than the number of tails

Accepted Answer

S = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT } => n(S) = 8 E = { HHH, HHT, HTH, THH } => n(E) = 4 P(E) = 4/8 = 1/2

Question 6

A number is selected from the numbers 1,2,3,4.......25.The probability for it to be divisible by 4 or 7 is:

Accepted Answer

Total numbers = 25 Numbers divisible by 4 or 7 are 4, 7, 8, 12, 14, 16, 20, 21, 24 = 9 The probability (divisible by 4 or 7) = 9/25

Question 7

Four persons are to be choosen from a group of 3 men, 2 women and 4 children. Find the probability of selecting 1 man,1 woman and 2 children.

Accepted Answer

Total number of persons = 9 Out of 9 persons 4 persons can be selected in 9 C 4 ways =126 1 man,1 woman and 2 children can be selected in 3 C 1 * 2 C 1 * 4 C 1 ways =36 So,required probability = 36/126 =2/7

Question 8

Two unbiased coin are tossed. What is the probability of getting atmost one head?

Accepted Answer

Here, S={HH,HT,TH,TT} Let E be the event of getting one head E={TT,HT,TH} P(E )= n(E)/n(S) =3/4

Question 9

You toss a coin AND roll a die. What is the probability of getting a tail and a 4 on the die?

Accepted Answer

Probability of getting a tail when a single coin is tossed =12 Probability of getting 4 when a die is thrown =16 Required probability =(12)×(16) = 1/12

Question 10

In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

Accepted Answer

250 numbers between 101 and 350 i.e. n(S)=250 n(E)=100th digits of 2 = 299−199 = 100 P(E)= n(E)/n(S) = 100/250 = 0.40

Question 11

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

Accepted Answer

5/12

Question 12

What is the probability of getting a sum 9 from two throws of a dice ?

Accepted Answer

In two throws of a die, n(s)=(6 x 6)=36 let E= Event of geting a sum 9={(3,6),(4,5),(5,4),(6,3)} P(E) = n(E)/n(S) = 4/36 = 1/9

Question 13

Let K and L be events on the same sample space, with P (K) = 0.8 and P (B) = 0.6. Are these two events are disjoint ?

Accepted Answer

These two events cannot be disjoint because P(K) + P(L) > 1. P(AꓴB) = P(A) + P(B) - P(AꓵB). An event is disjoint if P(A ꓵ B) = 0. If K and L are disjoint P(K ꓴ L) = 0.8 + 0.6 = 1.4 And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

Question 14

A fair six-sided die is rolled twice. What is the probability of getting 4 on the first roll and not getting 6 on the second roll ?

Accepted Answer

The two events mentioned are independent. The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied. P(getting first 4) = 1/6 P(no second 6) = 5/6 Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36

Question 15

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

Accepted Answer

S = { 1, 2, 3, 4, 5, 6 } => n(S) = 6 E = { 3, 6} => n(E) = 2 Therefore, P(E) = 2/6 =1/3

Question 16

The probabiltiy that throw of two dice yields a total of 5 or 6 is:

Accepted Answer

Total number of cases=36 Number of favourable cases(sum 5 or 6)=10 P(getting total of 5 or 6)=10/36=5/18

Question 17

The probability that a card drawn at random from the pack of playing cards may be either a queen or an ace is:

Accepted Answer

Number of queen cards = 4 Number of ace cards = 4 P(either a queen or ace) = 4/52 + 4/52= 8/52 = 2/13

Question 18

What is the probability that a couple with four children have atleast one girl?

Accepted Answer

Here,n = 4(children) P(girl)= 0.5 P(of atleast one girl)= 1 - P(no girls) = 1 - 0.0625 = 0.9375

Question 19

Find the probability of selecting exactly 2 children when four persons are choosen at random from a group of 3 men, 2 woman and 4 children.

Accepted Answer

4 persons can be selected from 9 in 9 C 4 ways =126 Fvaourable events = 4 C 2 * 5 C 2 =60 So,required probability = 60/126 = 10/21

Question 20

One card is drawn from deck of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the ball drawn is red.

Accepted Answer

Total number of elementary events = 52 There are 26 red cards,out of which one red card can be drawn in 26 C 1 ways =26. So,required probability = 26/52 = 1/2

Probability Questions