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…

Let us define following events
$B_1$ : Event of selection of bag $1$ ( $5$ Red, $5$ Green)
$B_2$ : Event of selection of bag $2$ ( $3$ Red, $5$ Green)
$B_3$ : Event of selection of bag $3$ ( $5$ Red, $3$ Green)
$R=$ Event of selection of Red ball
$G=$ Event of selection of Green ball
Option $\left(C\right)$ :
$P\left(\frac{G}{B_3}\right)=\frac{3}{8}$ (as bag $3$ has been selected now total balls in bag $3=8,$ green ball $=3$ )
Option $\left(\mathrm{B}\right)$ :
$P\left(G\right)=P\left(B_1\right)P\left(\frac{G}{B_1}\right)+P\left(B_2\right)P\left(\frac{G}{B_2}\right)+P\left(B_3\right)P\left(\frac{G}{B_3}\right)$
$=\frac{3}{10}\times \frac{5}{10}+\frac{3}{10}\times \frac{5}{8}+\frac{4}{10}\times \frac{3}{8}$
$=\frac{39}{80}$
Option $\left(D\right)$ :
$P\left(\frac{B_3}{G}\right)=\frac{P\left(B_3\cap G\right)}{P\left(G\right)}$
$=\frac{P\left(B_3\right)P\left(\frac{G}{B_3}\right)}{P\left(G\right)}$
$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\times \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}{\raisebox{1ex}{$35$}\!\left/ \!\raisebox{-1ex}{$80$}\right.} \left(\therefore P\left(G\right)=\frac{39}{80}\right)$
$=\frac{4}{13}$
Option $\left(A\right)$ :
$P\left(B_3\cap G\right)=P\left(B_3\right)P\left(\frac{G}{B_3}\right)$
$P\left(B_3\cap G\right)=P\left(B_3\right)P\left(\frac{G}{B_3}\right)$
$=\frac{3}{20}$