Question

A bag contains 4 black, 2 white and 6 red balls. Another bag contains 3 black and 5 white balls. An unbiased die is thrown. If either 1 or 2 appears, a ball is chosen from the first bag, otherwise a ball from the second bag is chosen. If the drawn ball is black then the probability that 2 appeared on the die is

• A. $\frac{2}{13}$
• B. $\frac{11}{13}$
• C. $\frac{6}{13}$
• D. $\frac{7}{13}$

Answer 1

The correct option is A $\frac{2}{13}$

Let the events be
$E_1$: Event of 1 appearing.
$E_2$: event of 2 appearing.
$E_3:\overline{E_1\cup E_2}=\overline{E_1}\cap \overline{E_2}$
B: Event of drawing a black ball.

Now
$P\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{1}{6},P\left(E_3\right)=\frac{4}{6}$

and $P\left(B/E_1\right)=\frac{4}{12}$

$P\left(B/E_2\right)=\frac{4}{12}$

$P\left(B/E_3\right)=\frac{3}{8}$

By Bayes' theorem:

$P\left(E_2/B\right)=\frac{P\left(E_2\right)P\left(B/E_2\right)}{P\left(E_1\right)P\left(B/E_1\right)+P\left(E_2\right)P\left(B/E_2\right)+P\left(E_3\right)P\left(B/E_3\right)}$

$=\frac{\frac{1}{6}\times \frac{4}{12}}{\frac{1}{6}\times \frac{4}{12}+\frac{1}{6}\times \frac{4}{12}+\frac{4}{6}\times \frac{3}{8}}$

$=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{3}+\frac{3}{2}}=\frac{1}{3}\times \frac{6}{13}=\frac{2}{13}$