Question

One bag contains $3$ red $5$ black balls. Another bag contains $6$ red $4$ 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

Probability of drawing a red ball from first bag $P\left(R\right)=\frac{3}{8}$

Probability of drawing a black ball from first bag $P\left(B\right)=\frac{5}{8}$

Now if one ball taken out from bag $1$ is of red colour ,then probability of drawing a red ball from bag $2$, is $P\left(R^{\prime }\right)=\frac{7}{11}$

Now if one ball taken out from bag $1$ is of black colour ,then probability

of drawing a red ball from bag $2$ is $P\left(R^{\prime \prime }\right)=\frac{6}{11}$

Total probability of getting a red ball from second bag

$=P\left(R\right)P\left(R^{\prime }\right)+P\left(B\right)P\left(R^{\prime \prime }\right)$

$=\frac{3}{8}\times \frac{7}{11}+\frac{5}{8}\times \frac{6}{11}=\frac{21+30}{88}=\frac{51}{88}$