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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 of them is black?
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Solution
Let B1 ≡ the event that the first ball is black
B2 ≡ the event that the second ball is black
N1 ≡ the event that the first ball is not black
N2 ≡ the event that the second ball is not black
Let A ≡ the event that at least one is black
A will happen if anyone of B1 ∩ N2, N1 ∩ B2, B1 ∩ B2 occurs.
These events are mutually exclusive
∴ P(A) = P(B1 ∩ N2) + P(N1 ∩ B2) + P(B1 ∩ B2)
= `"P"("B"_1)*"P"("N"_2//"B"_1) + "P"("N"_1)*"P"("B"_2//"N"_1) + "P"("B"_1)*"P"("B"_2/"B"_1)` ...(1)
Now, P(B1) = `4/15`, P(N1) = `11/15`
Since first ball is not replaced, the bag contains 3 black balls and 11 non-black balls.
`"P"("N"_2//"B"_1)` = probability that second ball is not black given that first black ball is not replaced
= `11/14`
Similarly, `"P"("B"_2//"N"_1) = 4/14, "P"("B"_2//"B"_1) = 3/14`
∴ from (1), we get,
P(A) = `4/15*11/14 + 11/15 * 4/14 + 4/15*3/14`
= `100/(15 xx 14)`
= `10/21`
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