Q. 55.0( 3 Votes )

A die is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on first two tosses.

Answer :

When a die is thrown three times the number of all favorable outcomes are n(S) = 216


Let E be the event that 4 occurs on the third toss, then the favorable outcomes will be


E ={(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)}


n(E) = 36


So the probability that 4 appears on the third toss is



Let F be the event that 6 and 5 appear on first two tosses respectively, and then the favorable outcomes will be


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


n(F) = 6


So the probability that 6 and 5 appear on first two tosses respectively will be



So the favorable outcomes that 6 and 5 appear respectively on first two tosses and 4 appears on the third toss will be


(EF) = {(6, 5, 4)}, n(EF) = 1


And the corresponding probability will be



So the conditional probability that if it is given that 6 and 5 appear respectively on first two tosses and 4 appears on the third toss will be



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