Given that all three faces are different in a throw of three dice, find the probability that
the sum is 9.

Explanation

Let A be the event thet the sum is 9.
`therefore A cap B`={(1,3,5),(1,5,3),(3,5,1),(3,1,5),(5,1,3),(5,3,1),(1,2,6),(1,6,2),(2,1,6),(2,6,1),(6,2,1),(6,1,2),(2,3,4),(2,4,3),(3,2,4),(2,3,4),(3,4,2),(4,3,2) }
`therefore absA capB=18`
`therefore P(A | B )=(P(A cap B))/(P(B))=(18//216)/(20//216)=9/10`