Probability

Sponsor Area

Question
CBSEENMA11015562

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

  • 1/50 

Solution

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)

Sponsor Area

Question
CBSEENMA11015596

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

  • 60

Solution

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

Question
CBSEENMA11015599

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

  • 2/3

Solution

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

Question
CBSEENMA11015611

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

  • 1124

Solution

C.

385

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