In a certain town, 60% of the families own a car, 30% own a house

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 Multiple Choice QuestionsMultiple Choice Questions

621.

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

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622.

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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623.

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

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  • 7/20

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624.

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is

  • At least 750 but less than 1000

  • At least 1000

  • Less then 500

  • At least 500 but less than 750


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625.

The probability that a non-leap year selected at random will have 53 Sunday is,

  • 0

  • 1/7

  • 2/7

  • 3/7


626.

Out of 7 constants and 4 vowels, words are formed each having 3 constants 2 vowels. The number of such words that can be formed is

  • 210

  • 25200

  • 2520

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627.

Let A and B be two events such that PA  B = 16, P(A  B) = 3145 and P(B) = 710 then

  • A and B are independent

  • A and B are mutually exclusive

  • PAB < 16

  • PBA < 16


628.

If Cr - 1n = 36, Crn = 84 and Cr  + 1n = 126, then the value of C8n is

  • 10

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  • 9

  • 8


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629.

In a group of 14 males and 6 females. 8 and 3 of the males and females, respectively are aged above 40 yr. The probability that a person selected at random from the group is aged above 40 yr given that the selected person is a female, is

  • 27

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630.

In a certain town, 60% of the families own a car, 30% own a house and 20% own both car and house. If a family is randomly chosen, then what is the probability that this family owns a car or a house but not both?

  • 0.5

  • 0.7

  • 0.1

  • 0.9


A.

0.5

Let A be the set of families who own a car and B be the set of families who own a house.

Then, P(A) = 60%, P(B) = 30%

and P(A  B) = 20%Now, P(A  B) = P(A) + P(B) - P(A  B)                            = 60 + 30 - 20 = 70 %

Now, families who owns a car or a house but not both

= A  B P(A  B) = P(A  B) - P(A  B)

                  = 70 - 20 = 50 % = 0.5


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