Question

Bag $A$ contains $1$ white, $2$ blue and $3$ red balls. Bag $B$ contains $3$ white, $3$ blue and $2$ red balls. Bag $C$ contains $2$ white, $3$ blue and $4$ red balls. One bag is selected at random and then two balls are drawn from the selected bag. Find the probability that the balls drawn are white and red.

Answer 1

Probability of selecting a bag out of three $=\frac{1}{3}$

Probability of getting a red and white ball from bag $A$ $=\frac{{}^1{C}_1{\times }^3{C}_1}{{}^6{C}_2}=\frac{3}{15}=\frac{1}{5}$

Probability of selecting a red and a white ball from bag $B$ $=\frac{{}^2{C}_1{\times }^3{C}_1}{{}^8{C}_2}=\frac{6}{28}=\frac{3}{14}$

Probability of selecting a red and a white ball from bag $C$ $=\frac{{}^4{C}_1{\times }^2{C}_1}{{}^9{C}_2}=\frac{8}{36}=\frac{2}{9}$

So the probability that balls are white and red $=\frac{1}{3}(\frac{1}{5}+\frac{3}{14}+\frac{2}{9})=\frac{1}{3}\left(\frac{126+135+140}{630}\right)=\frac{401}{1890}$