Question 1

Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ?

Accepted Answer

Given total students in the class = 60 Students who are taking Economics = 24 and Students who are taking Calculus = 32 Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4 Students who are taking calculus only = 32 - 4 = 28 probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

Question 2

In a simultaneous throw of two dice, what is the probability of getting a total of 7 ?

Accepted Answer

1/6

Question 3

A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of the heart is?

Accepted Answer

Here in this pack of cards, n(S) = 52 Let E = event of getting a queen of the club or a king of the heart Then, n(E) = 2 P(E) = n(E)/n(S) = 2/52 = 1/26

Question 4

Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is ?

Accepted Answer

n(S) = 20 n(Even no) = 10 = n(E) n(Prime no) = 8 = n(P) P(E U P) = 10/20 + 8/20 - 1/20 = 17/20

Question 5

A brother and a sister appear for an interview against two vacant posts in an office. The probability of the brother’s selection is 1/5 and that of the sister’s selection is 1/3. What is the probability that only one of them is selected?

Accepted Answer

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected) = 1 5 * 2 3 + 4 5 * 1 3 = 2 5

Question 6

A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:

Accepted Answer

Here, n(E) = 7 C 1 × 5 C 1 × 5 C 1 And, n(S) = 12 C 1 * 11 C 1 * 10 C 1 P(S) = 7 * 6 * 5 12 * 11 * 10 = 7/44

Question 7

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn find the probability the card drawn is red.

Accepted Answer

We know that: When one card is drawn from a pack of 52 cards The numbers of possible outcomes n(s) = 52 We know that there are 26 red cards in the pack of 52 cards ⇒ The numbers of favorable outcomes n(E) = 26 Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S) ∴ required probability = 26/52 = 1/2.

Question 8

A bag contains 8 red and 4 green balls. Find the probability that two balls are red and one ball is green when three balls are drawn at random.

Accepted Answer

n S = C 4 12 = 495 n E = C 2 8 × C 1 4 = 112 ∴ P ( E ) = 112 495

Question 9

If P(A)=4/9;then the odd against the event A is:

Accepted Answer

Here,P(A)=4/9=p/(p+q) P(odd against event A)=q/p=5/4

Question 10

If a card is drawn at random from a pack of 52 cards,what is the chance of getting a spade or ace?

Accepted Answer

Number of spades in a standard deck of cards=13 Number of aces in a standard deck of cards=4 And,one of the aces is a spade. So, 13 + 4 - 1 = 16 spades or aces to choose from. Therfore,probabiltiy of getting a spade or an ace=16/52=4/13

Question 11

One lady has 2 children, one of her child is boy, what is the probability of having both are boys ?

Accepted Answer

In a family with 2 children there are four possibilities: 1) the first child is a boy and the second child is a boy (bb) 2) the first child is a boy and the second child is a girl (bg) 3) the first child is a girl and the second child is a boy (gb) 4) the first child is a girl and the second child is a girl (gg) But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb). n(E)= both are boys=BB=1 n(S)= 3 Required probability P = n(E)/n(S) = 1/3.

Question 12

If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is ?

Accepted Answer

The number of exhaustive outcomes is 36. Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2 P(E) = 1 - 1/2 = 1/2.

Question 13

In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples ?

Accepted Answer

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ x ⁶C₁) As remaining person can be any one among three couples left. Required probability = (⁵C₂ x ⁶C₁)/¹⁰C₅ = (10 x 6)/252 = 5/21

Question 14

In a simultaneous throw of two dice, what is the probability of getting a doublet ?

Accepted Answer

In a simultaneous throw of two dice, n(S) = 6 x 6 = 36 Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} P ( E ) = n ( E ) n ( S ) = 6 36 = 1 6

Question 15

A group of five persons is formed from five boys and four girls. Find the probability that there are at least two girls in the group?

Accepted Answer

Total number of possible ways = 9 C 5 = 9 x 8 x 7 x 6 x 5 5 x 4 x 3 x 2 x 1 = 126 Number of favorable cases = 4 C 2 x 5 C 3 + 4 C 3 x 5 C 2 + 4 C 4 x 5 C 1 4 x 3 2 x 1 x 5 x 4 x 3 3 x 2 x 1 + 4 x 3 x 2 3 x 2 x 1 x 5 x 4 2 x 1 + 1 x 5 = 6 x 10 + 4 x 10 + 5 = 60 + 40 + 5 = 105 Therefore, required probability = 105/126 = 5/6

Question 16

Two dice are rolled simultaneously. Find the probability of getting a total of 9.

Accepted Answer

S = { (1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (2, 1), (2, 2),.........(6, 5), (6, 6) } => n(S) = 6 x 6 = 36 E = {(6, 3), (5, 4), (4, 5), (3, 6) } => n(E) = 4 Therefore, P(E) = 4/36 = 1/9

Question 17

Four dice are thrown simultaneously. Find the probability that two of them show the same face and remaining two show the different faces.

Accepted Answer

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in C 1 6 , ways. Now select two distinct number out of remaining 5 numbers which can be done in C 2 5 ways. Thus these 4 numbers can be arranged in 4!/2! ways. So, the number of ways in which two dice show the same face and the remaining two show different faces is C 1 6 × C 2 5 × 4 ! 2 ! = 720 => n(E) = 720 ∴ P E = 720 6 4 = 5 9

Question 18

A box contains 5 detective and 15 non-detective bulbs. Two bulbs are chosen at random. Find the probability that both the bulbs are non-defective.

Accepted Answer

n(S) = C 2 20 = 190 n(E) = C 2 15 = 105 Therefore, P(E) = 105/190 = 21/38

Question 19

The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?

Accepted Answer

Vowels are A I A I O, C A S T I G A T I O N (O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O) So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows : n ( E ) = 5 P 5 2 ! * 2 ! ( A , I a r e 2 t i m e s ) * 6 P 6 2 ! ( T i s 2 t i m e s ) = 21600 n ( S ) = 11 ! 2 ! * 2 ! * 2 ! ( A , I a r e 2 t i m e s ) = 4989600 21600 4989600 = 0 . 043

Question 20

In a throw of a coin, find the probability of getting a head.

Accepted Answer

Here, s={H,T} and E={H} P(E) = n(E)/n(S) = 1/2

Probability Questions