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
$\Rightarrow P\left(\frac{G}{A}\right)=\frac{\left( \, \right)^{4} C_{1}}{\left( \, \right)^{7} C_{1}}=\frac{4}{7} \, and \, P\left(\frac{G}{B}\right)=\frac{\left( \, \right)^{3} C_{1}}{\left( \, \right)^{7} C_{1}}=\frac{3}{7}$
Now, $P\left(\frac{B}{G}\right)=\frac{P \left(\right. B \left.\right) . P \left(\frac{G}{B}\right)}{P \left(\right. A \left.\right) P \left(\frac{G}{A}\right) + P \left(\right. B \left.\right) P \left(\frac{G}{B}\right)}$
$P\left(\frac{B}{G}\right)=\frac{\frac{1}{2} . \frac{3}{7}}{\frac{1}{2} . \frac{4}{7} + \frac{1}{2} . \frac{3}{7}}=\frac{3}{7}$