Question

The contents of three urns are as given below:
$1$ white, $2$ black, $3$ red balls
$2$ white, $1$ black, $1$ red balls
$4$ white, $5$ black, $3$ red balls
One urn is chosen at random and two balls are drawn. These happen to be white and red. What is the probability that they came from urn $2$

Answer 1

$\begin{array}{c}I\Rightarrow 1white2black3red\\ II\Rightarrow 2white1black1red\\ III\Rightarrow 4white5black3red\\ Let{E}_1,E_2,E_{3}and{}^{\prime }{A}^{\prime }denotethefollowingevent:\\ E_1=UrnIischoosen\\ E_2=UrnIIischoosen\\ E_3=UrnIIIischoosen\\ A=Twoballsaredrawnatrandom\&whiteandred.\\ ProbabilityofUrn-\\ P\left(E_1\right)=p\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}\\ case\left(I\right)choosingUrn\left(I\right)\\ Totalnumberofballs=6\\ andhavetochoose=2balls\left(1\right)\\ Numberofwhiteball=1\\ Numberofblackball=2\\ Numberofredball=3\\ P\left(A/E_1\right)=\frac{{}^1{c}_1{.}^3{c}_1}{{}^6{c}_2}\left[{}^n{c}_r=\frac{|!n}{|!n.|!n-n}\right]\\ =\frac{1\times 3}{15}=\frac{1}{5}\\ Similarly,P\left(A/E_2\right)=\frac{{}^2{c}_1{.}^1{c}_1}{{}^4{c}_2}\\ =\frac{2\times 1}{6}=\frac{1}{3}\\ P\left(A/E_3\right)=\frac{{}^4{c}_1{.}^3{c}_1}{{}^{12}{c}_2}\\ =\frac{4\times 3}{66}=\frac{2}{11}\\ usingBa{y}^{\prime }stheorem.\\ P\left(E_2/A\right)=\frac{P\left(E_2\right).P\left(A/E_2\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)+P\left(E_3\right)P\left(A/E_3\right)}\\ \frac{\frac{1}{3}\times \frac{1}{3}}{\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{11}}\\ \frac{\frac{1}{9}}{\frac{1}{15}+\frac{1}{9}+\frac{2}{33}}\\ \frac{\frac{1}{9}}{\frac{1}{3}\left(\frac{1}{3}+\frac{1}{5}+\frac{2}{11}\right)}\\ \frac{\frac{1}{3}}{\frac{\left(55+33+30\right)}{165}}\\ \frac{\frac{1}{3}}{\frac{118}{165}}\\ \frac{1}{3}\times \frac{165}{118}\\ P\left(E_2/A\right)=\frac{55}{118}Ans.\end{array}$