# Probability – Quiz 7

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# Probability – Quiz 7

### Introduction

Probability is one of important topic in Quantitative Aptitude Section. In Probability – Quiz 7 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Probability - Quiz 7 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?
A. $\frac {52}{221}$ B. $\frac {55}{190}$ C. $\frac {55}{221}$ D. $\frac {19}{221}$

C
We have n(s) = $^{52}{C}_{2} 52 = 52 \times \frac {51}{2 \times 1} = 1326$
Let A = event of getting both black cards
B = event of getting both queens
A∩B = event of getting queen of black cards
n(A) = $\frac {52 \times 51}{2 \times 1} = ^{26}{C}_{2} = 325, n(B)= \frac{26 \times 25 }{2 \times 1}= \frac {4 \times 3}{2 \times 1} = 6 and n(A∩B) = ^{4}{C}_{2} = 1$
P(A) = $\frac {n(A)}{n(S)} = \frac {325}{1326}$
P(B) = $\frac {n(B)}{n(S)} = \frac {6}{1326}$
P(A∩B) = $\frac {n(A∩B)}{n(S)} = \frac {1}{1326}$
P(A∪B) = P(A) + P(B) - P(A∩B) = $\frac {(325 + 6 - 1)}{1326} = \frac {330}{1326} = \frac {55}{221}$

### Q2

Two dice are tossed. The probability that the total score is a prime number is:
A. $\frac {5}{12}$ B. $\frac {1}{6}$ C. $\frac {1}{2}$ D. $\frac {7}{9}$

A
Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.
Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }
n(E) = 15.
P(E) = $\frac {n(E)}{n(S)} = \frac {15}{36} = \frac {5}{12}$

### Q3

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
A. $\frac {2}{91}$ B. $\frac {1}{22}$ C. $\frac {3}{22}$ D. $\frac {2}{77}$

A
Let S be the sample space.
Then, n(S) = number of ways of drawing 3 balls out of 15 = $^{15}{C}_{3} = \frac {15 \times 14 \times 13}{3 \times 2 \times 1} = 455$
Let E = event of getting all the 3 red balls.
n(E) = $^{5}{C}_{3} = \frac {5 \times 4 }{3 \times 2 \times 1} = 10$
=> P(E) = $\frac {n(E)}{n(S)} \frac {10}{455} = \frac {2}{91}$

### Q4

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
A. $\frac {2}{7}$ B. $\frac {5}{7}$ C. $\frac {1}{5}$ D. $\frac {1}{2}$

A
Total number of outcomes possible, n(S) = 10 + 25 = 35
Total number of prizes, n(E) = 10
P(E) = $\frac {n(E)}{n(s)} = \frac {10}{35} = \frac {2}{7}$

### Q5

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
A. $\frac {21}{46}$ B. $\frac {1}{5}$ C. $\frac {3}{25}$ D. $\frac {1}{50}$

A
Let, S - sample space E - event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25
= $^{25}{C}_{3}$
= 2300.
n(E) = $^{10}{C}_{1}$ × $^{15}{C}_{2}$ = 1050.
P(E) = $\frac {n(E)}{n(s)} = \frac {1050}{2300} = \frac {21}{46}$

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