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Given that all three faces are different in a throw of three dice, find the probability that the sum is 9. |
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Answer» <P> Solution :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` |
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