Question

Bag I contains 3 red and 4 black balls, Bag II contains 5 red and 2 black balls. Two balls are transferred at random from Bag I to Bag II and then a ball is drawn at random from Bag II. Find the probability that the drawn ball is red in colour.

Answer 1

Let, E₁, E₂, E₃ and A be event 

E₁ = Both transferred ball from Bag I to bag II are red.

E₂ = Both transferred ball from bag I to bag II are black

E₃ = out of two transferred ball one is red and other is black.

Now, ${\displaystyle P\left(E_1\right)=\frac{{}^3{C}_1}{{}^7{C}_2}=\frac{1}{7}}$

${\displaystyle P\left(E_2\right)=\frac{{}^4{C}_2}{{}^7{C}_2}=\frac{2}{7}}$

${\displaystyle P\left(E_3\right)=\frac{{}^3{C}_1{\times }^4{C}_1}{{}^7{C}_2}=\frac{4}{7}}$

Required probability

${\displaystyle P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right)\times P\left(\frac{A}{E_2}\right)}{P\left(E_1\right)\cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right)}\times P\left(\frac{A}{E_2}\right)+P\left(E_3\right)\times P\left(\frac{A}{E_3}\right)}$

${\displaystyle P\left(\frac{A}{E_1}\right)=\frac{7}{9},P\left(\frac{A}{E_2}\right)=\frac{2}{9},P\left(\frac{A}{E_3}\right)=\frac{6}{9}}$

Now, ${\displaystyle \frac{\frac{2}{7}\times \frac{2}{9}}{\frac{1}{7}\times \frac{7}{9}+\frac{2}{7}\times \frac{2}{9}+\frac{4}{7}\times \frac{6}{9}}}$

= ${\displaystyle \frac{4}{35}}$