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A die is thrown three times and the sum ...

A die is thrown three times and the sum of the 3 numbers shown is 15. The probability that the first throw was a four, is

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We have to find the concitionsl probability of sum 15 when 4 appears first.
Let the event of getting sum of three thrown numbers as 15 be A and the event to find P(A/B).
Now, `P(A//B)=(P(AnnB))/(P(B))`
Event `(AnnB)` is "4 appears in first throw and sum of numbers is 15"
`AnnB={(4,5,6),(4,6,5)}`
`P(A//B)=2/36=1/18`
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