There are 4 white and 6 black balls in a bag. Two balls are drawn at random. Find the probability that both balls drawn are black.

Explanation

To solve the problem of finding the probability that both balls drawn from a bag containing 4 white and 6 black balls are black, we can follow these steps:

### Step 1: Determine the total number of balls
We have:
- 4 white balls
- 6 black balls

Total number of balls = 4 + 6 = 10 balls.

### Step 2: Determine the total number of ways to draw 2 balls
We can use the combination formula \( nCr \) to find the total number of ways to choose 2 balls from 10. The formula for combinations is given by:

\[ nCr = \frac{n!}{r!(n-r)!} \]

For our case, we need to calculate \( 10C2 \):

\[ 10C2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2! \cdot 8!} \]

Calculating this gives:

\[ 10C2 = \frac{10 \times 9}{2 \times 1} = 45 \]

### Step 3: Determine the number of favorable outcomes (drawing 2 black balls)
Next, we need to find the number of ways to choose 2 black balls from the 6 black balls available. We calculate \( 6C2 \):

\[ 6C2 = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} \]

Calculating this gives:

\[ 6C2 = \frac{6 \times 5}{2 \times 1} = 15 \]

### Step 4: Calculate the probability
Now that we have both the total number of outcomes and the favorable outcomes, we can find the probability \( P \) that both balls drawn are black:

\[ P(\text{both black}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6C2}{10C2} = \frac{15}{45} \]

Simplifying this fraction:

\[ P(\text{both black}) = \frac{1}{3} \]

### Final Answer
The probability that both balls drawn are black is \( \frac{1}{3} \).

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