Question

There are two bags, one of which contains 5 black and 4 white balls while the other contains 3 black and 6 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the first bag. But if it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a white ball.

• A. $\frac{1}{2}$
• B. $\frac{14}{27}$
• C. $\frac{4}{27}$
• D. $\frac{16}{27}$

Answer 1

The correct option is D $\frac{16}{27}$

Let E denotes the event of selecting white ball.
Let A denotes that the outcome on the die is either 1 or 3
and B denotes the outcome other than 1 and 3.
Then,
$P\left(A\right)=\frac{2}{6}=\frac{1}{3}P\left(B\right)=\frac{4}{6}=\frac{2}{3}P\left(E|A\right)=\frac{4}{9}P\left(E|B\right)=\frac{6}{9}P\left(E\right)=P\left(E|A\right)P\left(A\right)+P\left(E|B\right)P\left(B\right)\Rightarrow P\left(E\right)=\frac{4}{9}\times \frac{1}{3}+\frac{6}{9}\times \frac{2}{3}\Rightarrow P\left(E\right)=\frac{16}{27}$