# XAT Probability Questions [Important PDF]

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

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

**Question 5: **A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one

ball is drawn from each bag, and the probability that both are green.

a) 13/70

b) 1/4

c) 6/35

d) 8/35

e) None of these

**5) Answer (C)**

**Solution:**

Total balls in bag A = 4 + 6 = 10

Probability that ball is green = $\frac{4}{10}$

Total balls in bag B = 3 + 4 = 7

Probability that ball is green = $\frac{3}{7}$

=> Required probability = $\frac{4}{10} \times \frac{3}{7}$

= $\frac{6}{35}$

**Question 6: **A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one

ball is drawn from each bag, and the probability that both are green.

a) 13/70

b) 1/4

c) 6/35

d) 8/35

e) None of these

**6) Answer (C)**

**Solution:**

Total balls in bag A = 4 + 6 = 10

Probability that ball is green = $\frac{4}{10}$

Total balls in bag B = 3 + 4 = 7

Probability that ball is green = $\frac{3}{7}$

=> Required probability = $\frac{4}{10} \times \frac{3}{7}$

= $\frac{6}{35}$

**Question 7: **A bag contains 2 red, 3 green and 2 blue balls. 2 balls are to be drawn randomly. What is the probability that the balls drawn contain no blue ball ?

a) 5/7

b) 10/21

c) 2/7

d) 11/21

e) None of these

**7) Answer (B)**

**Solution:**

Total number of balls = 2 + 3 + 2 = 7

Total number of outcomes = Drawing 2 balls out of 7

= $C^7_2 = \frac{7 \times 6}{1 \times 2} = 21$

Favourable outcomes = Drawing 2 balls out of 5 (so that none is blue)

= $C^5_2 = \frac{5 \times 4}{1 \times 2} = 10$

=> Required probability = $\frac{10}{21}$

**Question 8: **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

**8) 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 9: **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

**9) 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 10: **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) $\frac{8}{15}$

**10) 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 11: **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) $\frac{8}{15}$

**11) 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 12: **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

**12) 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 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.