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Two unblased die are thrown. Find the pr...

Two unblased die are thrown. Find the probability that the sum is 8 or greater if 3 appears on the first die.

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The correct Answer is:
`therefore P(B//A) = (P(A nn B))/(P(A)) = (1)/((18)/((1)/(6))) = (1)/(underset3(cancel18))xx (cancel6)/(1) = 1/3`
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