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?

#### Solution

Let B_{1} ≡ the event that first ball is black

B_{2} ≡ the event that second ball is black

N_{1} ≡ the event that first ball is not black

N_{2} ≡ the event that second ball is not black

Let A ≡ the event that at least one is black

A will happen if anyone of B_{1} ∩ N_{2}, N_{1} ∩ B_{2}, B_{1} ∩ B_{2} occurs.

These events are mutually exclusive

∴ P(A) = P(B_{1} ∩ N_{2}) + P(N_{1} ∩ B_{2}) + P(B_{1} ∩ B_{2})

= `"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(B_{1}) = `4/15`, P(N_{1}) = `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`