# 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? - Mathematics and Statistics

Sum

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 B1 ≡ the event that first ball is black

B2 ≡ the event that second ball is black

N1 ≡ the event that first ball is not black

N2 ≡ the event that 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

Concept: Event and Its Types
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