Question

A bag contains $4$ balls. Two balls are drawn at random and are found to be white. What is the probability that all balls are white?

• A. $\frac{2}{5}$
• B. $\frac{3}{5}$
• C. $\frac{4}{5}$
• D. $\frac{1}{5}$

Answer 1

The Correct Option is B

Let $E_1$, $E_2$, $E_3$ and $A$ be the events defined as follows : $E_1 =$ the bag contains $2$ white balls and $2$ non-white balls, $E_2 =$ the bag contains $3$ white balls and $1$ non-white ball, $E_3 =$ the bag contains all four white balls and $A =$ two white balls have been drawn from the bag. As the bags are selected at random, Then, $P(E_1) =\frac{1}{3}$, $P(E_2) =\frac{1}{3}$, $P(E_3) = \frac{1}{3}$ $P(A|E_1) = $ probability of drawing $2$ white balls when $E_1$ has occurred i.e. the bag contains $2$ white and $2$ non-white balls $= \frac{^{2}C_{2}}{^{4}C_{2}} = \frac{1}{6}$ $P(A|E_2) = $ probability of drawing $2$ white balls when $E_2$ has occurred i.e. the bag contains $3$ white and $1$ non-white ball $= \frac{^{3}C_{2}}{^{4}C_{2}} =\frac{3}{6}= \frac{1}{2}$ $P(A |E_3) = $ probability of drawing $2$ white balls when $E_3$ has occurred i.e. the bag contains all four white balls $= \frac{^{4}C_{2}}{^{4}C_{2}} =1$ We want to find $P(E_3|A)$. By Bayes' theorem, we have $P\left(E_{3}|A\right) = \frac{P\left(E_{3}\right)P\left(A |E_{3}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)+P\left(E_{3}\right)P\left(A|E_{3}\right)}$ $= \frac{\frac{1}{3}\times 1}{\frac{1}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{1}{2}+\frac{1}{3}\times 1}$ $= \frac{1}{\frac{1}{6}+\frac{1}{2}+1} = \frac{6}{10} = \frac{3}{5}$