We need to find

$$ P(R_1 \mid R_2) $$

Using Bayes’ theorem,

$$ P(R_1 \mid R_2)=\frac{P(R_1)P(R_2 \mid R_1)}{P(R_1)P(R_2 \mid R_1)+P(B_1)P(R_2 \mid B_1)} $$

Initially, box $B_1$ has 3 red and 4 black balls.

So,

$$ P(R_1)=\frac{3}{7}, \qquad P(B_1)=\frac{4}{7} $$

If the ball drawn from $B_1$ is red, then 4 red balls are added to $B_2$.

So $B_2$ will have:

$$ 3+4=7 \text{ red balls} $$

and 4 black balls.

Total balls in $B_2$:

$$ 7+4=11 $$

Thus,

$$ P(R_2 \mid R_1)=\frac{7}{11} $$

If the ball drawn from $B_1$ is black, then 3 black balls are added to $B_2$.

So $B_2$ will have 3 red balls and

$$ 4+3=7 $$

black balls.

Total balls in $B_2$:

$$ 3+7=10 $$

Thus,

$$ P(R_2 \mid B_1)=\frac{3}{10} $$

Now substitute in the formula:

$$ P(R_1 \mid R_2)=\frac{\frac{3}{7}\cdot \frac{7}{11}}{\frac{3}{7}\cdot \frac{7}{11}+\frac{4}{7}\cdot \frac{3}{10}} $$

Simplify:

$$ P(R_1 \mid R_2)=\frac{\frac{3}{11}}{\frac{3}{11}+\frac{6}{35}} $$

Taking LCM in the denominator:

$$ \frac{3}{11}+\frac{6}{35} = \frac{105+66}{385} = \frac{171}{385} $$

So,

$$ P(R_1 \mid R_2)=\frac{3}{11}\times \frac{385}{171} $$

$$ P(R_1 \mid R_2)=\frac{105}{171} $$

$$ P(R_1 \mid R_2)=\frac{35}{57} $$

So the correct option is

$$ \boxed{\frac{35}{57}} $$

Option A.

Probability

Continue this paper