Question

Bag $A$ contains $4$ green and $3$ red balls and bag $B$ contains $4$ red and $3$ green balls. One bag is taken at random and a ball is drawn and noted to be green. The probability that it comes from bag $B$ , is

• A. $\frac{2}{7}$
• B. $\frac{2}{3}$
• C. $\frac{3}{7}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

Let $A$ is the event that ball is drawn from Bag $A$ , $B$ is the event that ball is drawn from Bag $B$ and $G$ be the event that a green ball is drawn
$\Rightarrow P\left(\frac{G}{A}\right)=\frac{{\left(\right)}^4{C}_1}{{\left(\right)}^7{C}_1}=\frac{4}{7}andP\left(\frac{G}{B}\right)=\frac{{\left(\right)}^3{C}_1}{{\left(\right)}^7{C}_1}=\frac{3}{7}$
Now, $P\left(\frac{B}{G}\right)=\frac{P(B).P\left(\frac{G}{B}\right)}{P(A)P\left(\frac{G}{A}\right)+P(B)P\left(\frac{G}{B}\right)}$
$P\left(\frac{B}{G}\right)=\frac{\frac{1}{2}.\frac{3}{7}}{\frac{1}{2}.\frac{4}{7}+\frac{1}{2}.\frac{3}{7}}=\frac{3}{7}$