# Probability – Quiz 3

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# Probability – Quiz 3

### Introduction

Probability is one of important topic in Quantitative Aptitude Section. In Probability – Quiz 3 article candidates can find questions with an answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Probability - Quiz 3 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

There are 2 positive integers x and y. What is the probability that x + y is odd?
A. 4 B. 10 C. $\frac{1}{2}$ D. 36

C
Here we can have four cases:
a). x is even, y is even
b). x is odd, y is even
c). x is even, y is odd
d). x is odd, y is odd
Out of these four cases, in case 'b' and 'c' the sum will be odd.
So required probability = $\frac{2}{4}$ = $\frac{1}{2}$

### Q2

The probability that a arrow fired from a point will hit the target is $\frac{1}{4}$. Three such arrows are fired simultaneously towards the target from that very point. What is the probability that the target will be hit?
A. 64 B. 23 C. 67 D. $\frac{37}{64}$

D
Probability of a arrow not hitting the target = $\frac{3}{4}$
Probability that none of the 3 arrow will hit the target: = ${(\frac{3}{4})}^{3}$
= $\frac{27}{64}$
Probability that the target will hit at least once:
= 1 - $\frac{27}{64}$
= $\frac{37}{64}$

### Q3

A and B play a game where each is asked to select a number from 1 to 5. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is:
A. $\frac{24}{25}$ B. 25 C. $\frac{20}{25}$ D. 23

C
Total number of ways in which both of them can select a number each:
=5 × 5
=25
Total number of ways in which both of them can select a same number so that they both can win:
= 5 ways [They both can select {(1,1), (2,2), (3,3), (4,4), (5,5)}]
Probability that they win the prize:
= $\frac {Favorable Cases}{Total Cases}$
= $\frac{5}{25}$
Probability that they do not win a prize:
= 1 − $\frac{5}{25}$
= $\frac{20}{25}$

### Q4

A 5-digit number is formed by the digits 1, 2, 3, 4 and 5 without repetition. What is the probability that the number formed is a multiple of 4?
A. 2 B. $\frac{1}{3}$ C. 4 D. $\frac{1}{5}$

D
Total number of 5 digits number = 5! = 120.
Now to be multiply of 4, the last 2 digits of the number have to be divisible by 4. i.e. they must be 12, 24, 32 or 52.
Corresponding to each of these ways there are 3! = 6, i.e. 6 ways of filling the remaining 3 places.
The required probability
= $\frac{4 \times 6}{120}$
= $\frac{1}{5}$

### Q5

A box contains 9 red toys, 7 green toys and 9 blue toys. Each ball is of a different size. The probability that the red ball being selected is the smallest red toy, is:
A. $\frac{1}{9}$ B. $\frac{2}{21}$ C. $\frac{1}{25}$ D. $\frac{6}{25}$

A
Since, there are 9 red toys and all of them are of different sizes, the probability of choosing the smallest among them is $\frac{1}{9}$