A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
.: Let ${E_1}$ : `first bag is selected' and ${E_2}$
: `second bag is selected', : Red ball is drawn, $P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = \cfrac{1}{2}$
$P$ (red ball is drawn when first bag is selected) $= P\left( {A|{E_1}} \right) = \cfrac{4}{8}$
$P$ (redball is drawn when second bag is selected) $= P\left( {A|{E_2}} \right) = \cfrac{2}{8}$
Hence, the required probability
$P\left( {{E_1}|A} \right) = \cfrac{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right)}}{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right) + P\left( {A|{E_2}} \right)P\left( {{E_2}} \right)}}$
$= \cfrac{{\cfrac{4}{8} \times \cfrac{1}{2}}}{{\cfrac{4}{8} \times \cfrac{1}{2} + \cfrac{2}{8} \times \cfrac{1}{2}}} = \cfrac{4}{{4 + 2}} = \cfrac{4}{6} = \cfrac{2}{3}$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Probability. Curated by Sachin Sharma. Free for all students.