Step-by-step Solution

1. Define the events:

   • Let $E_1$ be the event of selecting box I.
   • Let $E_2$ be the event of selecting box II.
   • Let $R$ be the event of drawing a red ball.

2. State the given probabilities:

   • Since one of the two boxes is selected at random, the probability of selecting each box is: $P(E_1) = {1 \over 2}$ $P(E_2) = {1 \over 2}$
   • The conditional probabilities of drawing a red ball from each box are:
     • From box I: $P(R|E_1) = {{{n_1}} \over {{n_1} + {n_2}}}$
     • From box II: $P(R|E_2) = {{{n_3}} \over {{n_3} + {n_4}}}$

3. Formulate the problem using Bayes' Theorem:

   • We are given that the ball drawn is red, and the probability that it came from box II is ${1 \over 3}$. This can be written as: $P(E_2|R) = {1 \over 3}$
   • According to Bayes' Theorem: $P(E_2|R) = {{P(R|E_2)P(E_2)} \over {P(R|E_1)P(E_1) + P(R|E_2)P(E_2)}}$

4. Substitute the probabilities and solve for the condition:

   • Substitute the known values into the formula: ${1 \over 3} = {{{{{n_3}} \over {{n_3} + {n_4}}} \cdot {1 \over 2}} \over {{{{n_1}} \over {{n_1} + {n_2}}} \cdot {1 \over 2} + {{{n_3}} \over {{n_3} + {n_4}}} \cdot {1 \over 2}}}$
   • Cancel out the ${1 \over 2}$ term from the numerator and denominator: ${1 \over 3} = {{{{{n_3}} \over {{n_3} + {n_4}}}} \over {{{{n_1}} \over {{n_1} + {n_2}}} + {{{n_3}} \over {{n_3} + {n_4}}}}}$
   • Let's simplify this by letting $p_1 = P(R|E_1) = {{{n_1}} \over {{n_1} + {n_2}}}$ and $p_2 = P(R|E_2) = {{{n_3}} \over {{n_3} + {n_4}}}$. The equation becomes: ${1 \over 3} = {{{p_2}} \over {{p_1} + {p_2}}}$
   • Cross-multiplying gives: ${p_1} + {p_2} = 3{p_2}$ ${p_1} = 2{p_2}$
   • This is the condition that must be satisfied by the values of $n_1, n_2, n_3,$ and $n_4$. In other words: ${{{n_1}} \over {{n_1} + {n_2}}} = 2 \cdot {{{n_3}} \over {{n_3} + {n_4}}}$

5. Check each option against the derived condition:

   • Option A: ${n_1} = 3,{n_2} = 3,{n_3} = 5,{n_4} = 15$

     • LHS: ${p_1} = {3 \over {3 + 3}} = {3 \over 6} = {1 \over 2}$
     • RHS: $2{p_2} = 2 \cdot {{5} \over {5 + 15}} = 2 \cdot {5 \over {20}} = 2 \cdot {1 \over 4} = {1 \over 2}$
     • Since LHS = RHS, option A is correct.

   • Option B: ${n_1} = 3,{n_2} = 6,{n_3} = 10,{n_4} = 50$

     • LHS: ${p_1} = {3 \over {3 + 6}} = {3 \over 9} = {1 \over 3}$
     • RHS: $2{p_2} = 2 \cdot {{10} \over {10 + 50}} = 2 \cdot {{10} \over {60}} = 2 \cdot {1 \over 6} = {1 \over 3}$
     • Since LHS = RHS, option B is correct.

   • Option C: ${n_1} = 8,{n_2} = 6,{n_3} = 5,{n_4} = 20$

     • LHS: ${p_1} = {8 \over {8 + 6}} = {8 \over {14}} = {4 \over 7}$
     • RHS: $2{p_2} = 2 \cdot {{5} \over {5 + 20}} = 2 \cdot {5 \over {25}} = 2 \cdot {1 \over 5} = {2 \over 5}$
     • Since ${4 \over 7} \neq {2 \over 5}$, option C is incorrect.

   • Option D: ${n_1} = 6,{n_2} = 12,{n_3} = 5,{n_4} = 20$

     • LHS: ${p_1} = {6 \over {6 + 12}} = {6 \over {18}} = {1 \over 3}$
     • RHS: $2{p_2} = 2 \cdot {{5} \over {5 + 20}} = 2 \cdot {5 \over {25}} = 2 \cdot {1 \over 5} = {2 \over 5}$
     • Since ${1 \over 3} \neq {2 \over 5}$, option D is incorrect.

Conclusion

The options that satisfy the condition ${p_1} = 2{p_2}$ are A and B.