Question

The contents of three urns are as follows:
Urn 1 : 7 white, 3 black balls, Urn 2 : 4 white, 6 black balls, and Urn 3 : 2 white, 8 black balls. One of these urns is chosen at random with probabilities 0.20, 0.60 and 0.20 respectively. From the chosen urn two balls are drawn at random without replacement. If both these balls are white, what is the probability that these came from urn 3?

Answer 1

Let E₁, E₂ and E₃ denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the two balls drawn are white.

\[\therefore P\left( E_1 \right) = \frac{20}{100}\]

\[ P\left( E_2 \right) = \frac{60}{100} \]

\[ P\left( E_3 \right) = \frac{20}{100}\]

\[\text{ Now } , \]

\[P\left( A/ E_1 \right) = \frac{{}^7 C_2}{{}^{10} C_2} = \frac{21}{45}\]

\[P\left( A/ E_2 \right) = \frac{{}^4 C_2}{{}^{10} C_2} = \frac{6}{45}\]

\[P\left( A/ E_3 \right) = \frac{{}^2 C_2}{{}^{10} C_2} = \frac{1}{45}\]

\[\text{ Using Bayes' theorem, we get } \]

\[\text{ Required probability } = P\left( E_3 /A \right) = \frac{P\left( E_3 \right)P\left( A/ E_3 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + + P\left( E_3 \right)P\left( A/ E_3 \right)}\]

\[ = \frac{\frac{20}{100} \times \frac{1}{45}}{\frac{20}{100} \times \frac{21}{45} + \frac{60}{100} \times \frac{6}{45} + \frac{20}{100} \times \frac{1}{45}}\]

\[ = \frac{1}{21 + 18 + 1} = \frac{1}{40}\]