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 Multiple Choice QuestionsMultiple Choice Questions

1.

If PA = 112, PB = 512 and PBA = 115, then P(A B) is equal to :

  • 89180

  • 90180

  • 91180

  • 92180


A.

89180

 PA = 112, PB = 512 and PBA = 115We know that  PBA = PA  BPA  115 = PA  B115     PA  B = 1180Also, PA  B = PA + PB - PA  B                        = 112 + 512 - 1180                        = 15 + 75 - 1180 = 89180


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:

  • 517

  • 417

  • 516

  • 518


D.

518

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

Required probability =

           C15C19 × C14C18 × C14C17 × C14C19 × C15C18 × C13C17= 59 × 48 × 47 + 49 × 58 × 37= 80 + 60504 = 140504= 518


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

straight sigma subscript straight x superscript 2 space equals space fraction numerator sum from space to space of straight d subscript straight i superscript 2 over denominator straight n end fraction(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. 
therefore straight V subscript straight A over straight V subscript straight B space equals space 1 space left parenthesis As space begin inline style sum from space to space of end style space straight d subscript straight i superscript 2 space is space same space in space both space the space cases right parenthesis
387 Views

5.

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

  • 27

  • 23

  • 38

  • 18


A.

27

PA  B' = P(A)+P(B') - P(A).P(B')      0.8 = 0.3 + P(B') - 0.3 P(B')     0.5 = P(B') 0.7  P(B') = 57   P(B) = 1 - 57 = 27


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(straight A with bar on top) =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(top enclose straight B) =1/4.
The probability of that they contradict each other

space equals space straight P left parenthesis straight A right parenthesis. space straight P left parenthesis top enclose straight B right parenthesis space plus space straight P left parenthesis top enclose straight A right parenthesis. space straight P left parenthesis straight B right parenthesis
space equals space 4 over 5 space straight x 1 fourth space plus space 1 fifth space straight x 3 over 4 space equals space 1 fifth space plus space 3 over 20
space equals space 7 over 20

141 Views

7.

Probability of getting positive integral roots of the equation x - n = 0 for the integer n, 1  n  40 is

  • 15

  • 110

  • 320

  • 120


C.

320

Given, x2 - n = 0

                x = ± n

 n = 1, 4, 9, 16, 25, 36

 Required probability = 640 = 320


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

rightwards double arrow space straight x over straight y space equals 4 over 1
and space fraction numerator straight x over denominator straight x plus straight y end fraction space equals space 4 over 5
therefore space percent sign space of space boys space equals space 80

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.