There are three bags B _ 1 , B _ 2 and B _ 3 . The bag B _ 1...

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 $( C )$ :
$P\left(\dfrac{G}{{B}_{3}}\right) = \dfrac{3}{8}$ (as bag $3$ has been selected now total balls in bag $3 = 8,$ green ball $= 3$ )
Option $( B )$ :
$P\left(G\right) = P\left({B}_{1}\right)P\left(\dfrac{G}{{B}_{1}}\right) + P\left({B}_{2}\right)P\left(\dfrac{G}{{B}_{2}}\right) + P\left({B}_{3}\right)P\left(\dfrac{G}{{B}_{3}}\right)$
$= \dfrac{3}{10} \times \dfrac{5}{10} + \dfrac{3}{10} \times \dfrac{5}{8} + \dfrac{4}{10} \times \dfrac{3}{8}$
$= \dfrac{39}{80}$
Option $( D )$ :
$P\left(\dfrac{{B}_{3}}{G}\right) = \dfrac{P\left({B}_{3} \cap G\right)}{P\left(G\right)}$
$= \dfrac{P\left({B}_{3}\right)P\left(\dfrac{G}{{B}_{3}}\right)}{P ( G )}$
$= \dfrac{\dfrac{4}{10} \times \dfrac{3}{8}}{\dfrac{35}{80}}\left(∴ P\left(G\right) = \dfrac{39}{80}\right)$
$= \dfrac{4}{13}$
Option $( A )$ :
$P\left({B}_{3} \cap G\right) = P\left({B}_{3}\right)P\left(\dfrac{G}{{B}_{3}}\right)$
$P\left({B}_{3} \cap G\right) = P\left({B}_{3}\right)P\left(\dfrac{G}{{B}_{3}}\right)$
$= \dfrac{3}{20}$