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Question

One die is thrown three times and the sum of the thrown numbers is 15. The probability for which number 4 appears in first throw


A

118

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B

136

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C

19

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D

13

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Solution

The correct option is A

118


Explanation for the correct option:

Given: One die is thrown three times and the sum of the thrown numbers is 15.

Let A be the event of getting a sum of 15 of the three thrown numbers and B be the event of getting 4 on the first thrown

Now, only 2 cases are possible when sum of numbers appearing on the dice is 15 and first number is 4 and they are 4,6,5 and 4,5,6

and possible cases when first thrown is 4 will be 36 as other two thrown have total possible cases 6×6

Thus, the required probability is:

PAB=n(AB)n(B)=236[n(AB)=2beacusethereareonlytwocases]=118.

Hence the correct answer is option(A).


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