﻿ The number of ways of arranging 8 men and 4 women around a circular table such that no two women can sit together is | Probability

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# Probability

#### Multiple Choice Questions

1.

A fair coin is tossed-at a fixed number of times. If the probability of getting exactly 3 heads equals the probability of getting exactly 5 heads, then the probability of getting exactly one head is

• 1/64

• 1/32

• 1/16

• 1/8

2.

Let A and B be two events with . Then,  is equal to

• $\frac{1}{4}$

• $\frac{1}{3}$

• $\frac{1}{2}$

• $\frac{2}{3}$

3.

One ticket is selected at random from 100 tickets numbered 00, 01, 02,..., 98, 99. If x1 and x2 denote the sum and product of the digits on the tickets, then P(x1 = 9/x2 = 0) is equal to

• $\frac{1}{50}$

• $\frac{2}{19}$

• $\frac{19}{100}$

• None of these

4.

Two small squares on a chess board are chosen at random. Probability that they have a common side is

• $\frac{1}{3}$

• $\frac{1}{9}$

• $\frac{1}{15}$

• $\frac{1}{18}$

5.

Two dice are rolled simultaneously. The probability that the sum of the two numbers on the dice is a prime number, is

• $\frac{5}{12}$

• $\frac{7}{12}$

• $\frac{9}{12}$

• 0.25

6.

The events A andB have probabilities 0.25 and 0.50, respectively. The probability that both A and B occur simultaneously is 0.14, then the probability that neither A nor B occurs, is

• 0.39

• 0.29

• 0.11

• 0.25

7.

If  the coefficient of x is in expansion of  is 270, then k is equal to

• 1

• 2

• 3

• 4

8.

An unbiased coin is tossed to get 2 points forturning up a head and one point for the tail.If three unbiased coins are tossed simultaneously, then the probability of getting a total of odd number of points is

• $\frac{1}{2}$

• $\frac{1}{4}$

• $\frac{1}{8}$

• $\frac{3}{8}$

9.

Six faces of an unbiased die are numbered with 2, 3, 5, 7, 11 and 13. If two such dice are thrown, then the probability that the sum on the upper most faces of the dice is an odd number is

• $\frac{5}{18}$

• $\frac{5}{36}$

• $\frac{13}{18}$

• $\frac{25}{36}$

# 10.The number of ways of arranging 8 men and 4 women around a circular table such that no two women can sit together is8! 4! 8!4! $7!{}_{8}\mathrm{P}_{4}$

D.

$7!{}_{8}\mathrm{P}_{4}$

The number of ways of arranging 8 men = 7!

The number of ways of arranging 4 women such that no two women can sit together = 8P4

:. Required number of ways = 7!8P4