First slide

Theorems of probability

Question

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

Moderate

A

$\frac{1}{10}$
B

$\frac{3}{10}$
C

$\frac{3}{5}$
D

$\frac{1}{2}$

Solution

Let $A_{i} (i = 1, 2, 3, 4, 5)$ be the event that urn contains 2, 3, 4, 5 white balls and let B 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.

$∴ P (B / A_{2}) = \frac{2_{C_{2}}}{5_{C_{2}}} = \frac{1}{10}$

Similarly, $∴ P (B / A_{3}) = \frac{3_{C_{2}}}{5_{C_{2}}} = \frac{3}{10}$ ,

$∴ P (B / A_{4}) = \frac{4_{C_{2}}}{5_{C_{2}}} = \frac{3}{5}$ ,

$∴ P (B / A_{5}) = \frac{5_{C_{2}}}{5_{C_{2}}} = 1$

By baye’s theorem ,

$P (A_{5} / B) = \frac{P (A_{5}) P (B / A_{5})}{P (A_{2}) P (B / A_{2}) + P (A_{3}) P (B / A_{3}) + P (A_{4}) (B / A_{4}) + P (A_{5}) P (B / A_{5})}$

$= \frac{\frac{1}{4} .1}{\frac{1}{4} [\frac{1}{10} + \frac{3}{10} + \frac{3}{5} + 1]} = \frac{10}{20} = \frac{1}{2}$

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