Question 1026962: A fair die is rolled twice. Let X denote the sum of the two numbers resulting. Let E be the event that the first die roll resulted in a five. Calculate Pr (E|X=9), Pr(E|X=10) and Pr(E|X=11).
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! Let (r1, r2) be the ordered pair where r1 is the result of the first roll, and r2 is the number in the second roll.
Clearly X = r1 + r2.
For P(E|X = 9), there are four possible ordered pairs satisfying X = 9, namely
(6,3), (5,4), (4,5), and (3,6).
Of these, (5,4) corresponds to the event that the first roll is 5.
Thus, P(E|X = 9) = 1/4.
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For P(E|X = 10), there are three possible ordered pairs satisfying X = 10, namely
(6,4), (5,5), (4,6).
Of these, (5,5) corresponds to the event that the first roll is 5.
Thus, P(E|X = 10) = 1/3.
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For P(E|X = 11), there are only two possible ordered pairs satisfying X = 1, namely
(6,5), and (5,6).
Of these two, (5,6) corresponds to the event that the first roll is 5.
Thus, P(E|X = 11) = 1/2.
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