Q. 225.0( 2 Votes )

# <span lang="EN-US

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) =

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) =

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