An urn contains 4 balls. Two balls are drawn at random from the urn (without replacement) and are found to be white. What is the probability that all the four balls in the urn are white?

0.6

The correct answer is Option (1) → 0.6 ##

Let \(E_1\) : urn has 2 white balls 

\(E_2\) : urn has 3 white balls 

\(E_3\) : urn has 4 white balls 

\(A\) : 2 balls drawn are white

$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$

$P(A/E_1) = \frac{^{2}C_{2}}{^{4}C_{2}} = \frac{1}{6}$,

$P(A/E_2) = \frac{^{3}C_{2}}{^{4}C_{2}} = \frac{3}{6} = \frac{1}{2}$,

$P(A/E_3) = \frac{^{4}C_{2}}{^{4}C_{2}} = 1$

By Bayes' theorem,

$P(E_3/A) = \frac{P(E_3) \cdot P(A/E_3)}{P(E_1)P(A/E_1) + P(E_2)P(A/E_2) + P(E_3)P(A/E_3)}$

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

$= \frac{6}{10} = 0.6$