Question

One bag contains 3 red and 5 black balls. Another bag contains 6 red and 4 black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is red.

Answer 1

P(Red transferred and red drawn or black transferred and red drawn) $=\frac{3}{8}\times \frac{7}{11}+\frac{5}{8}\times \frac{6}{11}=\frac{51}{88}.$
Alternative: Let E: the ball drawn from second bag is red, $E_1$ : red ball is transferred from first bag and $E_2$: black ball is transferred from first bag.
$\therefore P\left(E_1\right)=\frac{3}{8},P\left(E_2\right)=\frac{5}{8},P\left(E|E_1\right)=\frac{7}{11},P\left(E|E_2\right)=\frac{6}{11}$
So, $P\left(E\right)=P\left(E_1\right)P\left(E|E_1\right)+P\left(E_2\right)P\left(E|E_2\right)=\frac{3}{8}\times \frac{7}{11}+\frac{5}{8}\times \frac{6}{11}=\frac{51}{88}$