# Probability – Quiz 8

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# Probability – Quiz 8

### Introduction

Probability is one of important topic in Quantitative Aptitude Section. In Probability – Quiz 8 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Probability - Quiz 8 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 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
A. $\frac {3}{7}$ B. $\frac {4}{7}$ C. $\frac {1}{8}$ D. $\frac {3}{4}$

B
Let number of balls = (6 + 8) = 14.
Number of white balls = 8.
P (drawing a white ball) = $\frac {8}{14}$ = $\frac {4}{7}$.

### Q2

One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
A. $\frac {3}{13}$ B. $\frac {1}{13}$ C. $\frac {3}{52}$ D. $\frac {9}{52}$

A
Clearly, there are 52 cards, out of which there are 12 face cards.
P (getting a face card) = $\frac {12}{52}$ = $\frac {3}{13}$.

### Q3

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:
A. $\frac {3}{20}$ B. $\frac {29}{34}$ C. $\frac {47}{100}$ D. $\frac {13}{102}$

D
Let S be the sample space.
Then, n(S) = $^{52}{C}_{2}$ = $\frac {(52 x 51)}{(2 x 1)}$ = 1326.
Let E = event of getting 1 spade and 1 heart.
n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = $^{13}{C}_{1} \times ^{13}{C}_{1} = 169$.
P(E) = $\frac {n(E)}{n(S)} = \frac {169}{1326} = \frac {13}{102}$.

### Q4

Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6.
A. $\frac {7}{18}$ B. $\frac {14}{35}$ C. $\frac {8}{18}$ D. $\frac {7}{35}$

A
Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = [(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)]
n(E) = 14.
Hence, P(E) = $\frac {n(E)}{n(S)} = \frac {14}{36} = \frac {7}{18}$

### Q5

What is the probability of getting 53 Mondays in a leap year?
A. $\frac {7}{35}$ B. $\frac {9}{35}$ C. $\frac {2}{7}$ D. 1

C
1 year = 365 days . A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 x 7 = 364 days
366 – 364 = 2 days
In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
1. Sunday, Monday
2. Monday, Tuesday
3. Tuesday, Wednesday
4. Wednesday, Thursday
5. Thursday, Friday
6. Friday, Saturday
7. Saturday, Sunday
Of these total 7 outcomes, the favorable outcomes are 2.
Hence the probability of getting 53 days = $\frac {2}{7}$

### Study Guide

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