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Question
Two dice are thrown simultaneously, If at least one of the dice show a number 5, what is the probability that sum of the numbers on two dice is 9?
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Solution
When two dice are thrown simultaneously, the sample space is
S = = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(S) = 36
Let A be the event that at least one die shows number 5.
∴ A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)}
∴ n(A) = 11
∴ P(A) = `("n"("A"))/("n"("S")) = 11/36`
Let B be the event that sum of the numbers on two dice is 9.
∴ B = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(B) = 4
∴ P(B) = `("n"("B"))/("n"("S")) = 4/36`
∴ A ∩ B is the event that one dice shows a 5 and the sum of the numbers is 9.
∴ A ∩ B = {(4, 5), (5, 4)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = `("n"("A" ∩ "B"))/("n"("S")) = 2/36 `
Now, probability that sum of the numbers on the dice is
9, given that one dice shows 5 is given by,
`"P"("B"/"A") = ("p"("A" ∩ "B"))/("P"("A")`
= `(2/36)/(11/36)`
= `2/11`
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