Problem 5

Explanation

Let:

• $P(h_1) =$ Probability of black in box 1
• $P(h_2) =$ Probability of black in box 2
• $P(p_1) =$ Probability of white in box 1
• $P(p_2) =$ Probability of white in box 2

Given the probability of getting black balls from both boxes is $\frac{3}{10}$. Since taking balls from box 1 and box 2 are independent events:

We need to find a pair of factors of $120$ where both are less than or equal to $20$ (maximum number of balls). The possible pair is $12$ and $10$ (or vice versa). Let $n(h_1) = 12$ and $n(h_2) = 10$.

Then the number of white balls is:

• Box 1: $20 - 12 = 8$ white balls
• Box 2: $20 - 10 = 10$ white balls

The probability that both balls are white is:

Given $\frac{m}{n} = \frac{1}{5}$. So $m = 1$ and $n = 5$. The value of $m + n = 1 + 5 = 6$.