Q. 105.0( 1 Vote )

A die is thrown three times. Find P(A/B) and P (B/A), if A=4 appears on the third toss, B=6 and 5 appear respectively on first two tosses.

Answer :

When a die is thrown 3 times, total possible outcomes are, 63 = 216


A= 4 appears on third toss =


{(1,1,4),(1,2,4),(1,3,4),(1,4,4),(1,5,4),(1,6,4)


(2,1,4),(2,2,4),(2,3,4),(2,4,4),(2,5,4),(2,6,4)


(3,1,4),(3,2,4),(3,3,4),(3,4,4),(3,5,4),(3,6,4)


(4,1,4), (4,2,4)(4,3,4),(4,4,4),(4,5,4),(4,6,4)


(5,1,4),(5,2,4),(5,3,4)(5,4,4),(5,5,4),(5,6,4)


(6,1,4),(6,2,4),(6,3,4),(6,4,4),(6,5,4),(6,6,4)}


=36


Hence P(A) =


B= 6,5 appears respectively on first two tosses


{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)}


= 6


Hence P(B) =


(A B) = 6,5 occurs on first two toss and 4 occur on third toss = (6,5,4) = 1


P (A B) =


Hence =


= = (answer)


= (answer)


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