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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?
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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