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,
$P\left(E_{1}\right)=\frac{6}{10}=\frac{3}{5}$ and
$P\left(E_{2}\right)=\frac{4}{10}=\frac{2}{5}$
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, $P\left(A \mid E_{1}\right)=\frac{6}{12}=\frac{1}{2}$
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 $P\left(A \mid E_{2}\right)=\frac{7}{12}$
By using law of total probability, we get
$P(A)=P\left(E_{1}\right) P\left(A \mid E_{1}\right)+P\left(E_{2}\right) P\left(A \mid E_{2}\right)$
$=\frac{3}{5} \times \frac{1}{2}+\frac{2}{5} \times \frac{7}{12}$
$=\frac{8}{15}$