Q. A 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) $\frac{1}{3}$
Explanation
(c)
$P\left( \frac{G}{A} \right) = \frac{\ ^{4}C_{1}}{\ ^{7}C_{1}} = \frac{4}{7}\ and\ P\left( \frac{G}{B} \right) = \frac{\ ^{3}C_{1}}{\ ^{7}C_{1}} = \frac{3}{7}$
Now, $P\left( \frac{B}{G} \right) = \frac{P(G).P\left( \frac{G}{B} \right)}{P(A)P\left( \frac{G}{A} \right) + P(B)P\left( \frac{G}{B} \right)}$
$= \frac{\frac{1}{2}.\frac{3}{7}}{\frac{1}{2}.\frac{4}{7} + \frac{1}{2}.\frac{3}{7}} = \frac{3}{7}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.