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?

There are three bags B₁, B₂, and B₃. The bag B₁ contains 5 red and 5 green balls, B₂ contains 3 red and 5 green balls, and B₃ contains 5 red and 3 green balls. Bags B₁, B₂ and B₃ have probabilities 3/10, 3/10 and 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?

$\text{Given: } B_1:\ 5R,\,5G;\quad B_2:\ 3R,\,5G;\quad B_3:\ 5R,\,3G$

$P(B_1)=\tfrac{3}{10},\quad P(B_2)=\tfrac{3}{10},\quad P(B_3)=\tfrac{4}{10}$

$\text{A bag is chosen first, then a ball is drawn from it.}$

$\text{To check: which of } (1),(2),(3),(4) \text{ is/are correct}$

$(1)\ P(B_3\cap G)=\tfrac{3}{10} \qquad (2)\ P(G\mid B_3)=\tfrac{3}{8}$

$(3)\ P(B_3\mid G)=\tfrac{5}{13} \qquad (4)\ P(G)=\tfrac{39}{80}$

$P(B_1)=\tfrac{3}{10},\quad P(B_2)=\tfrac{3}{10},\quad P(B_3)=\tfrac{4}{10}$

$P(G\mid B_1)=\tfrac{5}{10}=\tfrac{1}{2},\qquad P(R\mid B_1)=\tfrac{5}{10}=\tfrac{1}{2}$

$P(G\mid B_2)=\tfrac{5}{8},\qquad P(R\mid B_2)=\tfrac{3}{8}$

$P(G\mid B_3)=\tfrac{3}{8},\qquad P(R\mid B_3)=\tfrac{5}{8}$

$P(G\mid B_3)=\tfrac{3}{8}=\text{option }(2) \;\Rightarrow\; (2)\ \checkmark$

$P(B_3\cap G)=P(B_3)\cdot P(G\mid B_3)=\tfrac{4}{10}\times\tfrac{3}{8}$

$=\tfrac{12}{80}=\tfrac{3}{20}$

$\tfrac{3}{20} \neq \tfrac{3}{10} \;\Rightarrow\; (1)\ \times$

$P(G)=P(B_1)P(G\mid B_1)+P(B_2)P(G\mid B_2)+P(B_3)P(G\mid B_3)$

$=\tfrac{3}{10}\cdot\tfrac{1}{2}+\tfrac{3}{10}\cdot\tfrac{5}{8}+\tfrac{4}{10}\cdot\tfrac{3}{8}$

$=\tfrac{3}{20}+\tfrac{15}{80}+\tfrac{12}{80}$

$=\tfrac{12+15+12}{80}=\tfrac{39}{80}=\text{option }(4) \;\Rightarrow\; (4)\ \checkmark$

$P(B_3\mid G)=\dfrac{P(B_3\cap G)}{P(G)}=\dfrac{3/20}{39/80}$

$=\dfrac{3}{20}\times\dfrac{80}{39}=\dfrac{3\times 80}{20\times 39}$

$=\dfrac{12}{39}=\dfrac{4}{13}$

$\tfrac{4}{13} \neq \tfrac{5}{13} \;\Rightarrow\; (3)\ \times$

$(1)\ \times,\quad (2)\ \checkmark,\quad (3)\ \times,\quad (4)\ \checkmark$

$\boxed{\text{Correct options: }(2)\ \text{and}\ (4)}$