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Consider the three events:

A = rolling 5, 6, 8 or 9.

B = rolling 7

C = rolling any other number.

Let P be the probability of these events in one roll of a pair of dice.

Then P(A) = P(5) + P(6) + P(8) + P(9) = 18/36 = 1/2

P(B) = P(7) = 6/36 = 1/6

and P(C) = 1 - [P(A) + P(B)] = 1/3

Now P(A before B) = P(A or C followed by A before B)

= P(A) + P(C)*P(A before B)

= 1/2 + 1/3*P(A before B)

That is, P(A before B) = 1/2 + 1/3*P(A before B)

or 2/3*P(A before B) = 1/2

so that P(A before B) = 1/2*3/2 = 3/4

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Q: What is the probability of rolling either a 5 or 6 or 8 or 9 before rolling a 7 with two dice?
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