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 it is green. The probability that it comes from bag B is

The correct option is C

3/7

It is based on Baye's theorem.

probability of picked bag A $P\left(A\right)=\frac{1}{2}$

probability of picked bag B $P\left(B\right)=\frac{1}{2}$

probability of green ball picked from bag A

$=P\left(A\right).P\left(\frac{G}{A}\right)=\frac{1}{2}\times \frac{4}{7}=\frac{2}{7}$

probability of green ball picked from bag B

$=P\left(B\right).P\left(\frac{G}{B}\right)=\frac{1}{2}‘\times \frac{3}{7}=\frac{3}{14}$

total probability of green ball =$\frac{2}{7}+\frac{3}{14}=\frac{1}{2}$

probability of fact that green ball is drawn from bag B

$P\left(\frac{G}{B}\right)=\frac{P\left(B\right)P\left(\frac{G}{B}\right)}{P\left(A\right)P\left(\frac{G}{A}\right)+P\left(B\right)P\left(\frac{G}{B}\right)}=\frac{\frac{1}{2}\times \frac{3}{7}}{\frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{3}{7}}=\frac{3}{7}$