Question

The contents of urns I, II, III are as follows:
Urn I : 1 white, 2 black and 3 red balls
Urn II : 2 white, 1 black and 1 red balls
Urn III : 4 white, 5 black and 3 red balls.
One urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from Urns I, II, III?

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 and red.

$$\begin{gathered} \therefore P\left(E_1\right)=\frac{1}{3} \\ P\left(E_2\right)=\frac{1}{3} \\ P\left(E_3\right)=\frac{1}{3} \\ \mathrm{Now}, \\ P\left(A/E_1\right)=\frac{{}^1{C}_1{\times }^3{C}_1}{{}^6{C}_2}=\frac{3}{15}=\frac{1}{5} \\ P\left(A/E_2\right)=\frac{{}^2{C}_1{\times }^1{C}_1}{{}^4{C}_2}=\frac{2}{6}=\frac{1}{3} \\ P\left(A/E_3\right)=\frac{{}^4{C}_1{\times }^3{C}_1}{{}^{12}{C}_2}=\frac{12}{66}=\frac{2}{11} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\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{{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{5}}}{\frac{1}{3}\times \frac{1}{5}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{2}{11}}} \\ =\frac{{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{2}{11}}}=\frac{{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{33+55+30}{165}}}=\frac{33}{118} \\ \mathrm{Required}\mathrm{probability}=P\left(E_2/A\right)=\frac{P\left(E_2\right)P\left(A/E_2\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{{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{3}}}{\frac{1}{3}\times \frac{1}{5}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{2}{11}}} \\ =\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{2}{11}}}=\frac{{\displaystyle \frac{1}{3}}}{{\displaystyle \frac{33+55+30}{165}}}=\frac{55}{118} \\ \mathrm{Required}\mathrm{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{{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{2}{11}}}{\frac{1}{3}\times \frac{1}{5}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{2}{11}}} \\ =\frac{{\displaystyle \frac{2}{11}}}{{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{2}{11}}}=\frac{{\displaystyle \frac{2}{11}}}{{\displaystyle \frac{33+55+30}{165}}}=\frac{30}{118} \end{gathered}$$