Probability Questions for MAH-CET

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Question 1: In a box carrying one dozen of oranges one third have become bad.If 3 oranges taken out from the box random ,what is the probability that at least one orange out of the 3 oranges picked up is good ?

a) 1/55

b) 54/55

c) 45/55

d) 3/55

e) None of these

1) Answer (B)

Solution:

Total number of oranges in the box = 12

Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$

= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$

Number of oranges which became bad = $\frac{12}{3}=4$

Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$

Number of desired selection of oranges, n(E) = 220 – 4 = 216

$\therefore$ $P(E) = \frac{n(E)}{n(S)}$

= $\frac{216}{220}= \frac{54}{55}$

=> Ans – (B)

Instructions

Study the information carefully to answer the following questions:

A bucket contains 8 red, 3 blue and 5 green marbles.

Question 2: If 3 marbles are drawn at random, what is the probability that none is red ?

a) ${3 \over 8}$

b) ${1 \over {16}}$

c) ${1 \over {10}}$

d) ${3 \over {16}}$

e) None of these

2) Answer (C)

Solution:

Number of ways of drawing 3 marbles out of 16

$n(S) = C^{16}_3 = \frac{16 \times 15 \times 14}{1 \times 2 \times 3}$

= $560$

Out of the three drawn marbles, none is red, i.e., they will be either blue or green.

=> $n(E) = C^8_3 = \frac{8 \times 7 \times 6}{1 \times 2 \times 3}$

= $56$

$\therefore$ Required probability = $\frac{n(E)}{n(S)}$

= $\frac{56}{560} = \frac{1}{10}$

Question 3: If 2 marbles are drawn at random, what is the probability that both are green?

a) ${1 \over 8}$

b) ${5 \over {16}}$

c) ${2 \over 7}$

d) ${3 \over 8}$

e) None of these

3) Answer (E)

Solution:

Number of ways of drawing 2 marbles out of 16

$n(S) = C^{16}_2 = \frac{16 \times 15}{1 \times 2}$

= $120$

Out of the two drawn marbles, both are green

=> $n(E) = C^5_2 = \frac{5 \times 4}{1 \times 2}$

= $10$

$\therefore$ Required probability = $\frac{n(E)}{n(S)}$

= $\frac{10}{120} = \frac{1}{12}$

Question 4: If 4 marbles are drawn at random, what is the probability that 2 are red and 2 are blue ?

a) ${{11} \over {16}}$

b) ${3 \over {16}}$

c) ${11 \over {72}}$

d) ${3 \over {65}}$

e) None of these

4) Answer (D)

Solution:

Number of ways of drawing 4 marbles out of 16

=> $n(S) = C^{16}_4 = \frac{16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4}$

= $1820$

Out of the four drawn marbles, 2 are red and 2 are blue.

=> $n(E) = C^8_2 \times C^3_2 = \frac{8 \times 7}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}$

= $28 \times 3 = 84$

$\therefore$ Required probability = $\frac{n(E)}{n(S)}$

= $\frac{84}{1820} = \frac{3}{65}$

Instructions

Study the given information carefully and answer the questions that follow:
An urn contains 3 red, 6 blue, 2 green and 4 yellow marbles.

Question 5: If four marbles are picked at random, what is the probability that one is green, two are blue and one is red ?

a) 4/15

b) 17/280

c) 6/91

d) 11/15

e) None of these

5) Answer (C)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 4 marbles at random out of 15

=> $P(S) = C^{15}_4 = \frac{15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4}$

= $1365$

P(E) = Favorable outcomes

= Selecting 1 green, 2 blue and 1 red marble.

=> $P(E) = C^2_1 \times C^6_2 \times C^3_1$

= $2 \times \frac{6 \times 5}{1 \times 2} \times 3$

= $90$

$\therefore$ Required probability = $\frac{P(E)}{P(S)}$

= $\frac{90}{1365} = \frac{6}{91}$

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Question 6: If two marbles are picked at random, what is the probability that either both are red or both are green ?

a) 3/5

b) 4/105

c) 2/7

d) 5/91

e) None of these

6) Answer (B)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 2 marbles at random out of 15

=> $P(S) = C^{15}_2 = \frac{15 \times 14}{1 \times 2}$

= $105$

P(E) = Favorable outcomes

= Selecting 2 green or 2 red marbles.

=> $P(E) = C^2_2 + C^3_2$

= $1 + \frac{3 \times 2}{1 \times 2}$

= $4$

$\therefore$ Required probability = $\frac{P(E)}{P(S)}$

= $\frac{4}{105}$

Question 7: If four marbles are picked at random, what is the probability that at least one is yellow ?

a) 91/123

b) 69/91

c) 125/143

d) 1/3

e) None of these

7) Answer (B)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 4 marbles at random out of 15

=> $P(S) = ^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4}$

= $1365$

Let no yellow marble is selected.

P(E) = Favorable outcomes

= Selecting 4 out of 11 marbles.

=> $P(E) = ^{11}C_4$

= $\frac{11 \times 10 \times 9 \times 8}{1 \times 2 \times 3 \times 4}$

= $330$

$\therefore$ Required probability = $1 – \frac{P(E)}{P(S)}$

= $1 – \frac{330}{1365} = 1 – \frac{22}{91}$

= $\frac{91 – 22}{91} = \frac{69}{91}$

Question 8: If three marbles are picked at random, what is the probability that two are blue and one is yellow ?

a) 2/15

b) 6/91

c) 12/91

d) 3/15

e) None of these

8) Answer (C)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 3 marbles at random out of 15

=> $P(S) = ^{15} C_3 = \frac{15 \times 14 \times 13}{1 \times 2 \times 3}$

= $455$

P(E) = Favorable outcomes

= Selecting 2 blue and 1 yellow marble.

=> $P(E) =C^6_2 \times C^4_1$

= $\frac{6 \times 5}{1 \times 2} \times 4$

= $60$

$\therefore$ Required probability = $\frac{P(E)}{P(S)}$

= $\frac{60}{455} = \frac{12}{91}$

Question 9: If two marbles are picked at random, what is the probability that both are green ?

a) 2/15

b) 1/15

c) 2/7

d) 1

e) None of these

9) Answer (E)

Solution:

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 2 marbles at random out of 15

=> $P(S) = C^{15}_2 = \frac{15 \times 14}{1 \times 2}$

= $105$

P(E) = Favorable outcomes

= Selecting 2 green marbles.

=> $P(E) = C^2_2 = 1$

$\therefore$ Required probability = $\frac{P(E)}{P(S)}$

= $\frac{1}{105}$

Question 10: There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls ?

a) $\frac{191}{1547}$

b) $\frac{180}{1547}$

c) $\frac{280}{1547}$

d) $\frac{189}{1547}$

e) None of these

10) Answer (C)

Solution:

Total number of balls in the bag = 8 + 4 + 5 = 17

P(S) = Total possible outcomes

= Selecting 5 balls at random out of 17

=> $P(S) = C^{17}_5 = \frac{17 \times 16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4 \times 5}$

= $6188$

P(E) = Favorable outcomes

= Selecting 2 brown, 1 orange and 2 black balls.

=> $P(E) = C^8_2 \times C^4_1 \times C^5_2$

= $\frac{8 \times 7}{1 \times 2} \times 4 \times \frac{5 \times 4}{1 \times 2}$

= $28 \times 4 \times 10 = 1120$

$\therefore$ Required probability = $\frac{P(E)}{P(S)}$

= $\frac{1120}{6188} = \frac{280}{1547}$

Question 11: In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random. What is the probability that at least one ball is of green colour ?

a) $\frac{4}{5}$

b) $\frac{3}{5}$

c) $\frac{1}{5}$

d) $\frac{2}{5}$

e) None of these

11) Answer (D)

Solution:

There are 4 white, 4 red and 2 green balls and two balls are drawn at random.

Total possible outcomes = Selection of 2 balls out of 10 balls

= $C^{10}_2 = \frac{10 * 9}{1 * 2} = 45$

Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls

= $C^2_1 \times C^8_1 + C^2_2$

= 2*8 + 2 = 18

$\therefore$ Required probability = $\frac{18}{45} = \frac{2}{5}$

Question 12: A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked at random. What is the probability that the ball drawn is white ?

a) $\frac{7}{11}$

b) $\frac{7}{30}$

c) $\frac{5}{11}$

d) $\frac{7}{15}$

e) $\frac{8}{15}$

12) Answer (D)

Solution:

Probability of choosing bag 1 = (1/2)
Probability of choosing bag 2 = (1/2)
Probability of choosing white ball from bag 1 = 3/5
Probability of choosing white ball from bag 2 = 2/6
Probability of choosing bag 1 and white ball from it = (1/2)(3/5) = 3/10
Probability of choosing bag 2 and white ball from it = (1/2)(2/6) = 2/12
Probability of choosing a bag and drawing a white ball = (3/10) + (2/12) = (28/60) = (7/15)
Option D is the correct answer.

Question 13: In a bag, there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red ?

a) 29/35

b) 7/15

c) 23/35

d) 2/5

e) 19/35

13) Answer (C)

Solution:

Probability that at least 1 ball is red = 1 – probability that none of them is red.
Probability that none if the two balls is red = (9/15)(8/14)
Probability that at least 1 ball is red = 1 – probability that none of them is red. = 1- [(9/15)(8/14)] = (210-72)/210
= 138/210
=23/35
Option C is the correct answer.

Question 14: A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue in colour ?

a) $\frac{20}{91}$

b) $\frac{10}{91}$

c) $\frac{15}{91}$

d) $\frac{5}{91}$

e) $\frac{25}{91}$

14) Answer (C)

Solution:

Probability of drawing blue ball in first attempt = 5/15

Probability of drawing two green balls in the next two attempts = (6/14)(5/13)
Probability of drawing 2 green and 1 blue ball = (5/15)(6/14)(5/13) = 150/2730

Probability of drawing green ball in first attempt = 6/15
Probability of drawing blue ball in the next attempt = (5/14)
Probability of drawing green ball in the next attempt = (5/13)
Probability of drawing 2 red and 1 green ball = (6/15)(5/14)(5/13) = 150/2730

Probability of drawing two green balls in first two attempts = (6/15)(5/14)
Probability of drawing blue ball in the next attempt =(5/13)
Probability of drawing 2 red and 1 green ball = (6/15)(5/14)(5/13) = 150/2730

Probability of drawing 2 red balls and 1 green ball= 150/2730 + 150/2730 + 150/2730 = 3(150/2730) = 150/910 = 15/91
Option C is the correct answer

Question 15: In a sample , if a person is picked up randomly, the probability that the person is a smoker is $\frac{3}{5}$, and that of the person being male is $\frac{1}{2}$ .What is the probability that the person is both male as well as a smoker ?

a) $\frac{10}{11}$

b) $\frac{1}{5}$

c) $\frac{3}{5}$

d) Cannot be determined

e) None of these

15) Answer (D)

Solution:

Let’s assume the sample size is 100. Let the number of male smokers be x.

Total number of smokers = 3/5 * 100 = 60

Number of men = 1/2 * 100 = 50

Hence, number of male non-smokers is 50-x. Number of female smokers is 60-x and number of female non-smokers is x-10.

Hence, probability of a person picked at random being a smoker and a male = x/100

As we do not know the value of x, we cannot determine the probability. Hence, option D.

Question 16: Uma has three children, what is the probability that none of the children is a girl ?

a) $\frac{1}{2}$

b) $\frac{1}{16}$

c) $\frac{1}{3}$

d) $\frac{3}{4}$

e) None of these

16) Answer (E)

Solution:

The number of possible combinations is 2 * 2 * 2 = 8

The probability that none of the children is a girl is 1/8

Option e) is the correct answer.

Instructions

Answer the following questions based on the information given below.
A bowl contains 4 red, 3 green, 2 blue and 5 black marbles.

Question 17: If three marbles are drawn at random, what is the probability that at least one is red?

a) $\frac{2}{7}$

b) $\frac{4}{91}$

c) $\frac{61}{91}$

d) $\frac{2}{13}$

e) None of these

17) Answer (C)

Solution:

Probability that at least one marble is red = 1 – Probability that none of the balls are red

= 1 – $^{10}C_3 / ^{14}C_3$

= 1 – 720 / 2184 = 1 – 30/91 = 61 / 91

Question 18: If three marbles are drawn at random, what is the probability that none of them are black ?

a) $\frac{6}{13}$

b) $\frac{3}{91}$

c) $\frac{5}{91}$

d) $\frac{3}{13}$

e) None of these

18) Answer (D)

Solution:

The total number of marbles present is $5+4+3+2 = 14$
The number of ways of picking three marbles from them is $^{14}C_3 = 364$

If no marble is black, the three marbles should be picked from $4+3+2 = 9$ marbles.
The number of ways in which this can be done is $^9C_3 = 84$

Hence, the required probability is $\frac{84}{364} = \frac{21}{91} = \frac{3}{13}$

Question 19: If four marbles are drawn at random, what is the probability that two are red and two are blue ?

a) $\frac{6}{1001}$

b) $\frac{1}{143}$

c) $\frac{1}{13}$

d) $\frac{6}{91}$

e) None of these

19) Answer (A)

Solution:

Probability of drawing two red and two blue balls = $^4C_2 * ^2C_2 / ^{14}C_4$ = 6 / 1001

Option a) is the correct answer.

Question 20: If two marbles are drawn random, what is the probability that they are green ?

a) $\frac{4}{91}$

b) $\frac{51}{91}$

c) $\frac{4}{13}$

d) $\frac{1}{13}$

e) None of these

20) Answer (E)

Solution:

Probability of drawing 2 green = $^3C_2 / ^{14}C_2$ = 3 / (7*13) = 3 / 91

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