Question

Coloured balls are distributed in $3$ bags.
$B_1\to 1B,2W,3R$
$B_2\to 2B,4W,1R$
$B_3\to 4B,5W,5R$
A bag is selected at random and then $2$ balls are drawn from the selected bag. They happen to be black and red. What is the probability that the balls come from bag $1$?

Answer 1

$E1$ : Bag $I$ is selected

$E2$ : Bag $II$ is selected

Since events $E_1,E_2$ and $E_3$ are mutually exclusive and exhaustive events :

$\therefore P\left(E_1\right)=P\left(E_2\right)=P\left(E_3\right)=\frac{1}{3}$

$P\left(A/E_1\right)=$ Probability red and black balls are drawn from Bag $1$

$=\frac{{}^1{C}_1{\times }^3{C}_1}{{}^6{C}_2}=\frac{1\times 3}{\frac{6!}{2!4!}}=\frac{1\times 3\times 2}{6\times 5}=\frac{1}{5}$

$P\left(A/E_2\right)=$ Probability that red and black balls drawn from bag $II$

$=\frac{{}^2{C}_1{\times }^1{C}_1}{{}^7{C}_2}=\frac{2\times 1}{\frac{7!}{2!5!}}=\frac{2\times 2}{7\times 6}=\frac{2}{21}$

$P\left(A/E_3\right)=$ Probability that red and black balls drawn from bag $III$

$=\frac{{}^4{C}_1{\times }^5{C}_1}{{}^{14}{C}_2}=\frac{40}{\frac{14!}{2!12!}}=\frac{40}{91}$

Probability that two drawn balls which are black and red are from bag $1$.

$P\left(E_1/A\right)=\frac{P(E_1\times P(A/E_1\left)\right)}{\left[P\right(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3\left)\right]}$

$=\frac{\frac{1}{3}\times \frac{1}{5}}{(\frac{1}{3}\times \frac{1}{5})+(\frac{1}{3}\times \frac{2}{21})+(\frac{1}{3}\times \frac{20}{91})}$

$=\frac{273}{703}$