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Question
A bag contains 3 yellow and 5 brown balls. Another bag contains 4 yellow and 6 brown balls. If one ball is drawn from each bag, what is the probability that, both the balls are of the same color?
Solution
Let Y1 ≡ the event that yellow ball is drawn from first bag
B1 ≡ the event that brown ball is drawn from first bag
Y2 ≡ the event that yellow ball is drawn from second bag
B2 ≡ the event that brown ball is drawn from second bag
∴ P(Y1) = `3/8`, P(B1) = `5/8`
P(Y2) = `4/10`, P(B2) = `6/10`
Let A = event that both balls are of same colour i.e., both are yellow or both are brown.
∴ A = (Y1 ∩ Y2) ∪ (B1 ∩ B2)
Since events in the brackets are mutually exclusive
∴ P(A) = P(Y1 ∩ Y2) + P(B1 ∩ B2)
= P(Y1)·P(Y2) + P(B1)·P(B2) ...[events are independent]
= `3/8*4/10 + 5/8*6/10`
= `42/80`
= `21/40`.
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