Question

The bag $'A'$ contains $5$ white and $3$ black balls while the bag $'B'$ contains $4$ white and $7$ black balls. One of the bags is chosen at random and a ball is drawn from it. What is the probability that the ball is white?

• A. $\frac{87}{176}$
• B. $\frac{81}{172}$
• C. $\frac{82}{179}$
• D. $\frac{87}{179}$

Answer 1

Answer: (A)

$\frac{87}{176}$

Answer 2

Let $E_1$ be the event that bag $A$ is chosen, $E_2$ be the event that bag $B$ is chosen and $A$ be the event that white ball is drawn Note that $E_1$ and $E_2$ are mutually exclusive and exhaustive events. Since one of the bag is chosen at random, $P\left(E_{1}\right) = \frac{1}{2}$, $P\left(E_{2} \right) = \frac{1}{2}$ $P\left(A|E_{1}\right) =$ probability of drawing a white ball from bag $A =\frac{5}{8}$ $P\left(A|E_{2}\right) =$ probability of drawing a white ball from bag $B = \frac{4}{11}$ By using law of total probability, we get $P\left(A\right) = P\left(E_{1}\right) P\left(A|E_{1}\right) + P\left(E_{2}\right) P\left(A|E_{2}\right)$ $= \frac{1}{2}\cdot\frac{5}{8}+\frac{1}{2}\cdot\frac{4}{11}$ $= \frac{1}{2}\left(\frac{5}{8}+\frac{4}{11}\right)$ $= \frac{1}{2}\cdot\frac{87}{88} = \frac{87}{176}$