Question

A fair die is rolled. If $1$ turns up a ball is picked up at random from bag $P$. If $2$ or $3$ turns up, a ball is picked up from bag $Q$. If $4$, $5$ or $6$ turns up, a ball is picked up from bag $R$. Bag $P$ contains $5$ red and $3$ white balls; bag $Q$ contains $4$ red and $3$ white balls; bag $R$ contains $3$ red and $6$ white balls. The die is rolled, a bag is picked and a ball is drawn. Find the probability that a red ball is drawn.

• A. $\frac{154}{337}$
• B. $\frac{154}{339}$
• C. $\frac{155}{337}$
• D. $\frac{155}{336}$

Answer 1

Answer: (D)

Let $E_1$, $E_2$, $E_3$ and $A$ be the events defined as follows :
$E_1 =$ bag $P$ is picked up,
$E_2 =$ bag $Q$ is picked up,
$E_3 =$ bag $R$ is picked up
$A =$ a red ball is drawn
Note that $E_1$, $E_2$ and $E_3$ are mutually exclusive and exhaustive events.
Then, $P(E_1) = \frac{1}{6}$,
$P(E_2) = \frac{2}{6}$ and
$P(E_3) = \frac{3}{6}$
$P(A|E_1) =$ probability of drawing a red ball from bag $P = \frac{5}{8}$
$P(A|E_2) =\frac{4}{7}$ and $P(A|E_3) =\frac{3}{9}$
By using law of total probability, we get
$P(A) = P(E_1) P(A|E_1) + P(E_2) P(A|E_2) + P(E_3) P(A|E_3)$
$= \frac{1}{6}\cdot\frac{5}{8}+\frac{2}{6}\cdot\frac{4}{7}+\frac{3}{6}\cdot\frac{3}{9}$
$= \frac{5}{48}+\frac{4}{21}+\frac{1}{6} = \frac{155}{336}$