# Probability – Quiz 4

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# Probability – Quiz 4

### Introduction

Probability is one of important topic in Quantitative Aptitude Section. In Probability – Quiz 4 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 4 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

If five dice are thrown simultaneously, what is the probability of getting the sum as seven?
A. $\frac{15}{{(6)}^{5}}$ B. $\frac{11}{{(6)}^{5}}$ C. $\frac{10}{{(6)}^{5}}$ D. $\frac{5}{{(6)}^{5}}$

A
The following are two cases when the sum will be 7:
7 = 1 + 1 + 1 + 1 + 3 ⇒ $^{5}{C}_{1}$ ways = 5
7 = 1 + 1 + 1 + 2 + 2 ⇒ $^{5}{C}_{2}$ ways = 10
Total number of possible ways of throwing five dice = ${(6)}^{5}$
The required probability = $\frac{15}{{(6)}^{5}}$

### Q2

There are three events A, B and C, one of which must and only can happen. If the odds are 8:3 against A ,5:2 against B, the odds against C must be:
A. 13 : 7 B. 3 : 2 C. 43 : 34 D. 43 : 77

C
According to the question,
$\frac{P(A')}{P(A)}$ = $\frac{8}{3}$, P(A) = $\frac{3}{11}$ and P(A') = $\frac{8}{11}$
Also $\frac{P(B')}{P(B)}$ = $\frac{8}{3}$
P(B) = $\frac{2}{7}$ and P(B') = $\frac{8}{3}$
Now, out of A,B and C, one and only one can happen.
P(A) + P(B) + P(C) = 1
P(C) = $\frac{34}{77}$
P(C') = 1 − P(C)
= $\frac{43}{77}$
So odd against C
$\frac{P(C)}{P(C')}$ = $\frac{43}{34}$

### Q3

If the probability that X will live 15 year is 7/8 and that Y will live 15 years is 9/10, then what is the probability that both will live after 15 years?
A. $\frac{1}{20}$ B. $\frac{63}{80}$ C. $\frac{1}{5}$ D. None of these

B
Required probability
$\frac{7}{8} \times \frac{9}{10}$
$\frac{63}{80}$

### Q4

Four boys and three girls stand in queue for an interview. The probability that they stand in alternate positions is:
A. $\frac{1}{35}$ B. $\frac{1}{34}$ C. $\frac{1}{17}$ D. $\frac{1}{68}$

A
Total number of possible arrangements for 4 boys and 3 girls in a queue = 7!
When they occupy alternate position the arrangement would be like:
B G B G B G B
Thus, total number of possible arrangements for boys = (4× 3 × 2)
Total number of possible arrangements for girls = (3 × 2)
Required probability
= $\frac{4 \times 3 \times 2 \times 3 \times \times 2}{7!}$
= $\frac{1}{35}$

### Q5

A number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either 3 or 13?
A. $\frac{17}{30}$ B. $\frac{2}{5}$ C. $\frac{11}{30}$ D. $\frac{4}{15}$

B
We know that,
Probability = $\frac{Favorable Cases}{Total Cases}$
The probability that the number is a multiple of 3 is $\frac{10}{30}$.
Since favorable cases here {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
= 10 cases
Total cases = 30 cases
Similarly the probability that the number is a multiple of 13 is $\frac{2}{30}$.
Since favorable cases here {13, 26}
= 2 cases
Total cases = 30 cases
Neither 3 nor 13 has common multiple from 1 to 30.
Hence, these events are mutually exclusive events.
Therefore chance that the selected number is a multiple of 3 or 13 is:
= $\frac{10 + 2}{30}$
= $\frac{2}{5}$