JEE Advanced 2019 Paper 1Question 1
There are three bags $B_1$, $B_2$ and $B_3$. The bag $B_1$ contains 5 red and 5 green balls, $B_2$ contains 3 red and 5 green balls, and $B_3$ contains 5 red and 3 green balls. Bags $B_1$, $B_2$ and $B_3$ have probabilities $\frac{3}{10}$, $\frac{3}{10}$ and $\frac{4}{10}$ respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
Probability
Solution
$P(G|B_3)=\frac38$. $P(G)=\frac3{10}\cdot\frac12+\frac3{10}\cdot\frac58+\frac4{10}\cdot\frac38=\frac{39}{80}$. $P(B_3|G)=\frac{3/20}{39/80}=\frac4{13}$ and $P(B_3\cap G)=\frac3{20}$, so C, D false.