﻿ Engineering Entrance Exam Question and Answers | Probability - Zigya

## Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# Probability

#### Multiple Choice Questions

1.

If , then P(A $\cup$ B) is equal to :

• $\frac{89}{180}$

• $\frac{90}{180}$

• $\frac{91}{180}$

• $\frac{92}{180}$

A.

$\frac{89}{180}$

2.

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is

• 5040

• 6210

• 385

• 385

C.

385

10C1 + 10C2 + 10C3 + 10C4
= 10 + 45 + 120 + 210 = 385
124 Views

3.

A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box one at a time, the probability that they are alternatively either {odd, even, odd} or {even, odd, even}is:

• $\frac{5}{17}$

• $\frac{4}{17}$

• $\frac{5}{16}$

• $\frac{5}{18}$

D.

$\frac{5}{18}$

In out of 9 tickets, 5 tickets are odd numbers and 4 tickets are even number.

Required probability =

4.

Suppose a population A has 100 observations 101, 102, … , 200, and another population B has 100 observations 151, 152, … , 250. If VA and VB represent the variances of the two populations, respectively, then VA/VB is

• 1

• 9/4

• 4/9

• 4/9

A.

1 (Here deviations are taken from the mean) Since A and B both have 100 consecutive integers, therefore both have same standard deviation and hence the variance. 387 Views

5.

A and B are two independent events such that P(A $\cup$ B') = 0.8, and P(A) = 0.3. Then, P(B) is

• $\frac{2}{7}$

• $\frac{2}{3}$

• $\frac{3}{8}$

• $\frac{1}{8}$

A.

$\frac{2}{7}$

6.

The probability that A speaks truth is 4/ 5 , while this probability for B is 3/ 4 . The probability that they contradict each other when asked to speak on a fact is

• 3/20

• 1/20

• 7/20

• 7/20

C.

7/20

The probability of speaking truth of A, P(A) = 4/5.
The probability of not speaking truth of A, P( ) =1− 4./ 5 = 1/5.
The probability of speaking truth of B, P(B) = 3/4.
The probability of not speaking truth of B,P( ) =1/4.
The probability of that they contradict each other 141 Views

7.

Probability of getting positive integral roots of the equation x - n = 0 for the integer n,  is

• $\frac{1}{5}$

• $\frac{1}{10}$

• $\frac{3}{20}$

• $\frac{1}{20}$

C.

$\frac{3}{20}$

Given, x2 - n = 0

$\therefore$ n = 1, 4, 9, 16, 25, 36

8.

The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is

• 40

• 20

• 80

• 80

C.

80

52x + 42y = 50
(x + y) 2x = 8y 143 Views

9.

One ticket is selected at random from 50 tickets numbered 00, 01, 02, ... , 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals

•  1/14

• 1/7

• 5/14

• 5/14

A.

1/14

A = Events that sum of the digits on selected ticket is 8
= {08, 17, 26, 35, 44}
n(A) = 5
Event that product of digits is zero
= {00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20,30, 40}
⇒ n(B) = 14

=P(A/B) = (5/14)

168 Views

10.

A manufacturer of cotter pins knows that 5% of his product is defective. He sells pins in boxes of 100 and guarantees that not more than one pin will be defective in a box. In order to find the probability that a box will fail to meet the guaranteed quality, the probability distribution one has to employ is

• binomial

• poisson

• normal

• exponential

B.

poisson

Required probability distribution is poisson distribution.