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 Y_{1} ≡ the event that yellow ball is drawn from first bag

B_{1} ≡ the event that brown ball is drawn from first bag

Y_{2} ≡ the event that yellow ball is drawn from second bag

B_{2} ≡ the event that brown ball is drawn from second bag

∴ P(Y_{1}) = `3/8`, P(B_{1}) = `5/8`

P(Y_{2}) = `4/10`, P(B_{2}) = `6/10`

Let A = event that both balls are of same colour i.e., both are yellow or both are brown.

∴ A = (Y_{1} ∩ Y_{2}) ∪ (B_{1} ∩ B_{2})

Since events in the brackets are mutually exclusive

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

= P(Y_{1})·P(Y_{2}) + P(B_{1})·P(B_{2}) ...[events are independent]

= `3/8*4/10 + 5/8*6/10`

= `42/80`

= `21/40`.