# Probability – Quiz 2

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# Probability – Quiz 2

### Introduction

Probability is one of important topic in Quantitative Aptitude Section. In Probability – Quiz 2 article candidates can find a question with an answer. By solving this question candidates can improve and maintain, speed, and accuracy in the exams. Probability - Quiz 2 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 letters B, G, I, N and R are rearranged to form the word 'Bring'. Find its probability:
A. $\frac{1}{120}$ B. $\frac{1}{40}$ C. $\frac{1}{24}$ D. $\frac{1}{76}$

A
There are total 5 letters. The probability that B gets the first position is $\frac{1}{5}$
The probability that G is in the second position is $\frac{1}{4}$
Likewise, probability for I, N and G
Hence required probability:
= $\frac{1}{5}\times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times 1$
= $\frac{1}{120}$

### Q2

Dinesh speaks truth in 3/4 cases and Abhishek lies in 1/5 cases. What is the percentage of cases in which both Dinesh and Abhishek contradict each other in stating a fact?
A. 60% B. 35% C. 20% D. 15%

B
D and A will contradict each other when one speaks truth and other speaks lies.
Probability of D speak truth and A lies
= $\frac{3}{4} \times \frac{1}{5}$
$\frac{3}{20}$
Probability of A speak truth and D lies
= $\frac{4}{5} \times \frac{1}{4}$
= $\frac{1}{5}$
The two probabilities are mutually exclusive.
Hence, probabilities that D and A contradict each other:
= $\frac{3}{20} + \frac{1}{5}$
= $\frac{7}{20}$
= 0.35 $\times$ 100 %
= 35 %

### Q3

A set of cards bearing the number 200-299 is used in a game. if a card is drawn at random, what is the probability that it is divisible by 3:
A. 0.66 B. 0.33 C. 0.44 D. 0.55

B
Total numbers of cards = 100 {and not 99}
Multiples of three = 33
The required probability,
= $\frac{33}{100}$
= 0.33

### Q4

When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 11?
A. $\frac{5}{6}$ B. $\frac{11}{12}$ C. $\frac{1}{6}$ D. $\frac{1}{12}$

B
Instead of finding the probability of this event directly, we will find the probability of the non-occurrence of this event and subtract it from 1 to get the required probability.
Combination whose sum of 12 is (6,6)
Combinations whose sum of 11 is (5,6), (6,5).
Therefore, there are totally 3 occurrences out of 36 occurrences that satisfy the given condition.
Probability whose sum of two numbers is greater than or equal to 11:
= $\frac{3}{36} = \frac{1}{12}$
Hence probability whose sum of two numbers is lesser than 11:
= 1 - $\frac{1}{12}$
= $\frac{11}{12}$

### Q5

A dice is rolled three times and the sum of the numbers appearing on the uppermost face is 15. The chance that the first roll was a four is:
A. $\frac{2}{5}$ B. $\frac{1}{5}$ C. $\frac{1}{6}$ D. None of these

B
The sum of numbers can be 15 in the following three ways :
Case I: 15 = 3 + 6 + 6
The first, second and third throws can be (3, 6, 6), (6, 3, 6) and (6, 6, 3) respectively.
Total number of ways in which 3,6 and 6 can be obtained = 3
Case II: 15 = 4 + 5 + 6
The first, second and third throws can be 4, 5 and 6.
Total number of ways in which 4,5 and 6 can be obtained = 6
Case III: 15 = 5 + 5 + 5
The first, second and third throws can be 5, 5 and 5.
Total number of ways in which 5,5, and 5 can be obtained = 1.
Hence, The total number of ways = 3 + 6 + 1 = 10
The total number of ways in which the first roll will be 4 is 2.
Required chance =$\frac{2}{10}$ = $\frac{1}{5}$