Q. 225.0( 2 Votes )

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Answer :

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

E=No of outcomes for getting sum greater than or equal to 10


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


Therefore, P(E) =


Case 1:


F= No of outcomes for 5 appears on first die =


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


Hence P(B) =


(E F ) = outcomes of getting a sum greater than or equal to 10 and first die showing 5


= {(5,6),(5,5)= 2


Hence P(E F) =


Therefore, = (answer)


Case 2:


F = 5 appears on atleast one die =


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


P(F) =


(E F ) = outcomes of getting a sum greater than or equal to 10 and at least die showing 5


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


Hence P(E F) =


Therefore, = (answer)


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