MAH-CET Probability Questions PDF [Most Important with Solutions]

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Probability Questions PDF (1)
Probability Questions PDF (1)

Probability Questions for MAH-CET

Here you can download a free Probability questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Probability of answers for the given questions. These questions will help you to make practice and solve the Probability questions in the MAH MBA CET exam. Utilize this best PDF practice set which is included answers in detail. Click on the below link to download the Probability MCQ PDF for MBA-CET 2022 for free.

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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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