This problem involves finding the probability of drawing two balls of the same color from a box containing 6 white balls and 4 black balls.

Total number of balls

The total number of balls is: $6 \text{ (white)} + 4 \text{ (black)} = 10 \text{ balls}.$

Total ways to choose 2 balls

The total number of ways to choose 2 balls from 10 is calculated using the combination formula: $C(n, k) = \frac{n!}{k!(n-k)!}.$ Here: $C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \cdot 9}{2 \cdot 1} = 45.$

Ways to choose 2 balls of the same color

1. Two white balls: $C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6 \cdot 5}{2 \cdot 1} = 15.$

2. Two black balls: $C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4 \cdot 3}{2 \cdot 1} = 6.$

Total favorable outcomes

The total number of favorable outcomes (same color) is: $15 \text{ (white-white)} + 6 \text{ (black-black)} = 21.$

Probability

The probability of drawing two balls of the same color is: $P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{21}{45}.$

Simplify the fraction: $P = \frac{7}{15}.$

Final Answer:

The probability of drawing two balls of the same color is: $\boxed{\frac{7}{15}}$

Would you like a step-by-step explanation of how the formula works or further assistance with similar problems?

Here are five related questions for practice:

1. What is the probability of drawing two balls of different colors?
2. If one ball is replaced after each draw, how would the probabilities change?
3. What is the probability of drawing at least one white ball in two draws?
4. How does the probability change if there are 5 white and 5 black balls instead?
5. How can we calculate the probability of drawing three balls of the same color in a similar problem?

Tip: Always ensure the total number of combinations is calculated correctly when working with probability problems.