Question

Bag $P$ contains $6$ red and $4$ blue balls and bag $Q$ contains $5$ red and $6$ blue balls. $A$ ball is transferred from bag $P$ to bag $Q$ and then a ball is drawn from bag $Q$. What is the probability that the ball drawn is blue?

• A. $\frac{7}{15}$
• B. $\frac{8}{15}$
• C. $\frac{4}{19}$
• D. $\frac{8}{19}$

Answer 1

Answer: (B)

Let $E_1$, $E_2$ and $A$ be the events defined as follows:
$E_1 =$ red ball is transferred from bag $P$ to bag $Q$
$E_2 =$ blue ball is transferred from bag $P$ to bag $Q$
$A =$ the ball drawn from bag $Q$ is blue
As the bag $P$ contains $6$ red and $4$ blue balls,
$P(E_1) = \frac{6}{10} = \frac{3}{5}$ and
$P(E_2) = \frac{4}{10} = \frac{2}{5}$
Note that $E_1$ and $E_2$ are mutually exclusive and exhaustive events.
When $E_1$ has occurred i.e. a red ball has already been transferred from bag $P$ to $Q$, then bag $Q$ will contain $6$ red and $6$ blue balls,
So, $P(A|E_1) = \frac{6}{12} = \frac{1}{2}$
When $E_2$ has occurred i.e. a blue ball has already been transferred from bag $P$ to $Q$, then bag $Q$ contains $5$ red and $7$ blue balls,
So $P(A|E_2) = \frac{7}{12}$
By using law of total probability, we get
$P(A) = P(E_1) P(A|E_1) + P(E_2) P(A|E_2)$
$= \frac{3}{5} \times \frac{1}{2} + \frac{2}{5} \times \frac{7}{12}$
$=\frac{8}{15}$