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We know, when a pair of dice is thrown, total possible outcomes are = 36

A=No of outcomes for getting sum of 8 are

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

Therefore, P(A) =

B= No of outcomes for at least 5 has appeared once =

= {(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,5)} =11

Hence P(B) =

(A B) = outcomes of getting sum as 8 with at least one die showing 5

= {(3,5),(5,3)= 2

Hence P(A B)

Therefore, =

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