# Probability Quiz 9

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# Probability Quiz 9

### Introduction

Probability is one of important topic in Quantitative Aptitude Section. In Probability Quiz 9 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Probability Quiz 9 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

A bag contains 2 red caps, 4 blue caps, 3 yellow caps and 5 green caps. If three caps are picked at random, what is the probability that none is green?
A. $\frac {2}{13}$ B. $\frac {3}{13}$ C. $\frac {1}{13}$ D. $\frac {5}{13}$

B
Total caps = 14
Probability = $^{5}{C}_{0} \times \frac {^{9}{C}_{3}}{^{14}{C}_{3}} = \frac {3}{13}$

### Q2

A box contains 27 marbles some are blue and others are green. If a marble is drawn at random from the box, the probability that it is blue is 1/3. Then how many number of green marbles in the box?
A. 10 B. 15 C. 14 D. 18

D
Blue marble – x
$\frac {^{x}{C}_{1}}{^{27}{C}_{1}} = \frac {1}{3}$
$\frac {x}{27} = \frac {1}{3} \rightarrow x = \frac {27}{3} = 9$
No of green marbles = Total – Blue marble = 27 - 9 = 18

### Q3

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 red colour?
A. $\frac {4}{3}$ B. $\frac {7}{3}$ C. $\frac {1}{3}$ D. $\frac {2}{3}$

D
Total Balls = 10
Other than red ball = $^{6}{C}_{2}$
$\frac {^{6}{C}_{2}}{^{10}{C}_{2}} = \frac {1}{3} \rightarrow 1 - \frac {1}{3} = \frac {2}{3}$

### Q4

P and Q are sitting in a ring with 11 other persons. If the arrangement of 11 persons is at random, then the probability that there are exactly 4 persons between them?
A. $\frac {1}{3}$ B. $\frac {4}{7}$ C. $\frac {1}{5}$ D. $\frac {1}{6}$

D
Fix the position of P, then Q can be sit in 12 positions, so total possible outcome = 12
Now, exactly 4 persons are sitting between them. This can be done in two ways as shown in figure, so favourable outcomes = 2
So, probability = $\frac {2}{12} = \frac {1}{6}$

### Q5

A six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6. Find the probability that the number is divisible by 4.
A. $\frac {3}{17}$ B. $\frac {4}{15}$ C. $\frac {4}{19}$ D. $\frac {4}{17}$

B
For a number to be divisible by 4, the last two digit should be divisible by 4.
So possible cases – 12, 16, 24, 32, 36, 52, 56, 64 (last two digits)
So favourable outcomes = 24 + 24 + 24 + 24 + 24 + 24 + 24 + 24 = 192
So p = $\frac {192}{720} = \frac {4}{15}$

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