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.
