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 3 years ago
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers. Can someone please take a look at my work and show me where I am going wrong?
 3 years ago
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers. Can someone please take a look at my work and show me where I am going wrong?

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bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1okay, so this is what I have: P(A)  probability that at least 1 is 6 P(B)  probability that the die land on two diff numbers

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1There are the following options for at least one die to be 6: {(1,6);(2,6),(3,6),(4,6),(5,6),(6,6),(6,5),(6,4),(6,3),(6,2)(6,1)}

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1P(B) = 1  P(both land on the same number) = 1  (6/36) = 1(1/6) = 5/6

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1since there are 6 ways in which we can have two same numbers.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1P(AB) = P(A U B)/P(B)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1P(A U B) is they are different and at least one die is 6, so we just take the options above: {(1,6);(2,6),(3,6),(4,6),(5,6),{{(6,6)}},(6,5),(6,4),(6,3),(6,2)(6,1)}, but take away (6,6) So P(A U B) = 10/36

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1So now we have: P(AB) = (10/36)/(5/6) = (10/36)*(6/5) = (2/6) = 1/3

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1OMG.. i got it right lol

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1sorry for bothering you guys, but i was being dumb earlier

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.1your notation is not correct...should be... \[P(AB)=\frac{P(A\cap B)}{P(B)}\]

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1yea P(AintersectB), not P(AUB)
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