# Probability – Quiz 5

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# Probability – Quiz 5

### Introduction

Probability is one of important topic in Quantitative Aptitude Section. In Probability – Quiz 5 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Probability - Quiz 5 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

The probability that a number selected at random from the first 50 natural numbers is a composite number is
A. $\frac {21}{25}$ B. $\frac {4}{25}$ C. $\frac {17}{25}$ D. $\frac {8}{25}$

C
The number of exhaustive events = $^{50}{C}_{1}$ = 50.
We have 15 primes from 1 to 50.
Number of favorable cases are 34.
Required probability = $\frac {34}{50}$ = $\frac {17}{25}$.

### Q2

Coin is tossed Five times. What is the probability that there is at the least one tail?
A. $\frac {31}{32}$ B. $\frac {1}{16}$ C. $\frac {1}{2}$ D. $\frac {1}{32}$

A
Let P(T) be the probability of getting least one tail when the coin is tossed five times.
$P \bar{(T)}$ = There is not even a single tail.
i.e. all the outcomes are heads.
$P \bar{(T)}$ = $\frac {1}{32}$ ; P(T) = 1 - $\frac {1}{32}$ = $\frac {31}{32}$

### Q3

If a number is chosen at random from the set {1, 2, 3, ...., 100}, then the probability that the chosen number is a perfect cube is -
A. $\frac {1}{25}$ B. $\frac {1}{2}$ C. $\frac {4}{13}$ D. $\frac {1}{10}$

A
We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
$\frac {4}{100}$ = $\frac {1}{25}$.

### Q4

Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is
A. $\frac {1}{2}$ B. $\frac {16}{19}$ C. $\frac {4}{5}$ D. $\frac {17}{20}$

D
n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(EᴜP) = $\frac {10}{20}$ + $\frac {8}{20}$ - $\frac {1}{20}$ = $\frac {17}{20}$

### Q5

If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is -
A. $\frac {1}{4}$ B. $\frac {1}{2}$ C. $\frac {3}{4}$ D. $\frac {3}{5}$

B
The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = $\frac {18}{36}$ = $\frac {1}{2}$
P(E) = 1 - $\frac {1}{2}$ = $\frac {1}{2}$.