Suppose there are two boxes $B_1$ and $B_2$, each having 3 red and 4 black balls. One ball is drawn at random from $B_1$. If it is red, 4 red balls are put into $B_2$, otherwise 3 black balls are put into $B_2$. Then one ball is randomly drawn from $B_2$. If this ball is red, what is the conditional probability that the ball drawn from $B_1$ was also red?

Let $E_1$ be the event that a red ball is drawn from $B_1$.

Let $E_2$ be the event that a black ball is drawn from $B_1$.

Let $A$ be the event that a red ball is drawn from $B_2$.

The initial probabilities are:

$P(E_1) = \dfrac{3}{7}$

$P(E_2) = \dfrac{4}{7}$

If $E_1$ occurs, 4 red balls are added to $B_2$. The number of red balls in $B_2$ becomes $3 + 4 = 7$ and black balls remain 4. Total balls = 11.

$P(A|E_1) = \dfrac{7}{11}$

If $E_2$ occurs, 3 black balls are added to $B_2$. The number of red balls in $B_2$ remains 3 and black balls become $4 + 3 = 7$. Total balls = 10.

$P(A|E_2) = \dfrac{3}{10}$

By Bayes' theorem, the required probability is $P(E_1|A)$:

$P(E_1|A) = \dfrac{P(A|E_1)P(E_1)}{P(A|E_1)P(E_1) + P(A|E_2)P(E_2)}$

Substituting the values:

$P(E_1|A) = \dfrac{\dfrac{7}{11} \times \dfrac{3}{7}}{\dfrac{7}{11} \times \dfrac{3}{7} + \dfrac{3}{10} \times \dfrac{4}{7}}$

$P(E_1|A) = \dfrac{\dfrac{3}{11}}{\dfrac{3}{11} + \dfrac{12}{70}}$

$P(E_1|A) = \dfrac{\dfrac{3}{11}}{\dfrac{3}{11} + \dfrac{6}{35}}$

$P(E_1|A) = \dfrac{\dfrac{105}{385}}{\dfrac{105 + 66}{385}}$

$P(E_1|A) = \dfrac{105}{171} = \dfrac{35}{57}$

Answer : $\dfrac{35}{57}$