1. Given data

Let $G$ denote the event that the chosen ball is green.

The bags are chosen with probabilities: $P(B_1)=\frac{3}{10},\quad P(B_2)=\frac{3}{10},\quad P(B_3)=\frac{4}{10}$

Composition of bags:

• $B_1$: $5$ red, $5$ green $\Rightarrow P(G\mid B_1)=\frac{5}{10}=\frac12$
• $B_2$: $3$ red, $5$ green $\Rightarrow P(G\mid B_2)=\frac{5}{8}$
• $B_3$: $5$ red, $3$ green $\Rightarrow P(G\mid B_3)=\frac{3}{8}$

2. Check Option A

Option A states: $P(G\mid B_3)=\frac38$

From the composition of $B_3$, this is true.

So, A is correct.

3. Check Option C: total probability of getting a green ball

Using total probability: $P(G)=P(B_1)P(G\mid B_1)+P(B_2)P(G\mid B_2)+P(B_3)P(G\mid B_3)$

Substitute values: $P(G)=\frac{3}{10}\cdot\frac12+\frac{3}{10}\cdot\frac58+\frac{4}{10}\cdot\frac38$

Compute each term: $\frac{3}{10}\cdot\frac12=\frac{3}{20}$ $\frac{3}{10}\cdot\frac58=\frac{15}{80}=\frac{3}{16}$ $\frac{4}{10}\cdot\frac38=\frac{12}{80}=\frac{3}{20}$

So, $P(G)=\frac{3}{20}+\frac{3}{16}+\frac{3}{20}$

Take LCM $80$: $P(G)=\frac{12}{80}+\frac{15}{80}+\frac{12}{80}=\frac{39}{80}$

Hence, C is correct.

4. Check Option B: probability selected bag is $B_3$ given ball is green

Using Bayes' theorem: $P(B_3\mid G)=\frac{P(B_3\cap G)}{P(G)}$

Now, $P(B_3\cap G)=P(B_3)P(G\mid B_3)=\frac{4}{10}\cdot\frac38=\frac{12}{80}$

And from above, $P(G)=\frac{39}{80}$

Therefore, $P(B_3\mid G)=\frac{12/80}{39/80}=\frac{12}{39}=\frac{4}{13}$

But option B says $\frac{5}{13}$, which is incorrect.

So, B is incorrect.

5. Check Option D: probability that selected bag is $B_3$ and chosen ball is green

$P(B_3\cap G)=P(B_3)P(G\mid B_3)=\frac{4}{10}\cdot\frac38=\frac{12}{80}=\frac{3}{20}$

Option D says $\frac{3}{10}$, which is false.

So, D is incorrect.

6. Final conclusion

Correct options are: $\boxed{A,\ C}$

This matches the stored correct answer.