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Question
An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, What is the probability that at least one ball is black?
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Solution
Total number of balls in the urn = 4 + 5 + 6 = 15
Two balls can be drawn without replacement in 15C2 = `(15xx14)/(1xx2)` = 105 ways
∴ n(S) = 105
Let A be the event that at least one ball is black
i.e., 1 black and 1 non-black or 2 black and 0 non-black.
1 black ball can be drawn out of 4 black balls in 4C1 = 4 ways and 1 non-black ball can be drawn out of the remaining 11 non-black balls in 11C1 = 11 ways
∴ 1 black and 1 non black ball can be drawn in 4 × 11 = 44 ways
Also, 2 black balls can be drawn from 4 black balls in 4C2 = `(4xx3)/(1xx2)` = 6 ways
∴ n(A) = 44 + 6 = 50
∴ Required probability = P(A) = `("n"("A"))/("n"("S"))`
= `50/105`
= `10/21`
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