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.
When a die is thrown 3 times, total possible outcomes are, 63 = 216
A= 4 appears on third toss =
Hence P(A) =
B= 6,5 appears respectively on first two tosses
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) =
= = (answer)
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