A bag contains 15 red balls and 20 black balls. Each ball is numbered either 1 or 2 or 3. 20% of the red balls are numbered 1 and 40% of them are numbered 3. Similarly, among the black balls, 45% are numbered 2 and 30% are numbered 3. A boy picks a ball at random. He wins if the ball is red and numbered 3 or if it is black and numbered 1 or 2. What are the chances of his winning?

The correct option is $\frac{4}{7}$.

Let us break down the problem step-by-step to find the total probability of the boy winning.
First, let us determine the total number of balls in the bag:
Number of red balls = 15
Number of black balls = 20
Total number of balls = 15 + 20 = 35

Now, let us calculate the number of balls of each number (1, 2, or 3) for both red and black balls based on the given percentages.

For Red Balls (Total = 15):
- 20% of the red balls are numbered 1:
$15\times \frac{20}{100}=3$ red balls with number 1.
- 40% of the red balls are numbered 3:
$15��\frac{40}{100}=6$ red balls with number 3.
- The remaining red balls must be numbered 2:
$15-(3+6)=6$ red balls with number 2.

For Black Balls (Total = 20):
- 45% of the black balls are numbered 2:
$20\times \frac{45}{100}=9$ black balls with number 2.
- 30% of the black balls are numbered 3:
$20\times \frac{30}{100}=6$ black balls with number 3.
- The remaining black balls must be numbered 1:
$20-(9+6)=5$ black balls with number 1.

According to the winning conditions, the boy wins if:
1. The ball is red and numbered 3, OR
2. The ball is black and numbered 1 or 2.

Let us count the number of winning balls:
- Red balls numbered 3 = 6
- Black balls numbered 1 = 5
- Black balls numbered 2 = 9

Total winning balls = 6 + 5 + 9 = 20

The probability (chances) of his winning is the ratio of the number of winning balls to the total number of balls:
$\text{Probability}=\frac{20}{35}$

Simplifying the fraction by dividing the numerator and the denominator by their greatest common divisor, 5:
$\text{Probability}=\frac{4}{7}$