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Probability
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
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1/14
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1/7
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5/14
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1/50
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)
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
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40
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20
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80
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60
C.
80
52x + 42y = 50
(x + y) 2x = 8y
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
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1
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9/4
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4/9
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2/3
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. 
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
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5040
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6210
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385
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1124
C.
385
10C1 + 10C2 + 10C3 + 10C4= 10 + 45 + 120 + 210 = 385
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