Question

An urn contains five balls. Two balls are drawn and found to be white. The probability that all the balls are white is

• A. $\frac{1}{10}$
• B. $\frac{3}{10}$
• C. $\frac{3}{5}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Answer 2

Let $A_i ( i = 2, 3, 4, 5)$ be the event that urn contains $2, 3, 4, 5$ white balls and let B be the event that two white balls have been drawn then we have to find $P (A_5/B)$.
Since the four events $A_2, A_3, A_4$ and $A_5$ are equally likely
we have $P (A_2) = P (A_3) = P (A_4) = P(A_5) = \frac{1}{4}$
$P(B/A_2)$ is probability of event that the urn contains 2 white balls and both have been drawn.
$\therefore P\left(B/A_{2}\right) = \frac{^{2}C_{2}}{^{5}C_{2}} = \frac{1}{10}$
Similarly $P\left(B/A_{3}\right) = \frac{^{3}C_{2}}{^{5}C_{2}} = \frac{3}{10}$
$P\left(B/A_{4}\right) = \frac{^{4}C_{2}}{^{5}C_{2}} = \frac{3}{5}$
$P\left(B/A_{5}\right) = \frac{^{5}C_{2}}{^{5}C_{2}} = 1 .$
By Baye?s theorem,
$P\left(A_{5}/B\right) = \frac{P\left(A_{5} \right) P\left(B/A_{5}\right)}{P\left(A_{2}\right)P\left(B/A_{2}\right) + P\left(A_{3}\right)P\left(B/A_{3}\right)+P\left(A_{4}\right)\left(B/A_{4}\right)+P\left(A_{5}\right)P\left(B/A_{5}\right)}$
$= \frac{\frac{1}{4}.1}{\frac{1}{4} \left[ \frac{1}{10} + \frac{3}{10}+ \frac{3}{5} +1\right]} = \frac{10}{20}= \frac{1}{2}$