A bag contains 4 white, 5 red and 6 black balls. Three are drawn at random. Find the probability that (i) no ball drawn is black, (ii) exactly 2 are black (iii) all are of the same colour.

Answer

To solve the problem, we need to find the probabilities for three different scenarios involving the drawing of balls from a bag containing 4 white, 5 red, and 6 black balls. The total number of balls in the bag is:

Total balls = 4 (white) + 5 (red) + 6 (black) = 15 balls.

We will denote the total number of ways to choose 3 balls from 15 as \( \binom{15}{3} \).

### Step 1: Calculate the total number of ways to choose 3 balls from 15.

\[ \text{Total ways} = \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455. \]

### Step 2: Find the probability that no ball drawn is black.

To find this probability, we need to calculate the number of ways to choose 3 balls from the 9 non-black balls (4 white + 5 red).

\[ \text{Ways to choose 3 non-black balls} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84. \]

Now, the probability that no ball drawn is black is:

\[ P(\text{no black}) = \frac{\text{Ways to choose 3 non-black}}{\text{Total ways}} = \frac{84}{455}. \]

### Step 3: Find the probability that exactly 2 balls drawn are black.

To find this probability, we need to choose 2 black balls from the 6 available and 1 non-black ball from the 9 available.

\[ \text{Ways to choose 2 black balls} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15. \]
\[ \text{Ways to choose 1 non-black ball} = \binom{9}{1} = 9. \]

Thus, the total number of ways to choose exactly 2 black balls and 1 non-black ball is:

\[ \text{Total ways for exactly 2 black} = \binom{6}{2} \times \binom{9}{1} = 15 \times 9 = 135. \]

Now, the probability that exactly 2 balls drawn are black is:

\[ P(\text{exactly 2 black}) = \frac{135}{455}. \]

### Step 4: Find the probability that all balls drawn are of the same color.

We need to consider three cases: all white, all red, and all black.

1. **All white:**
\[ \text{Ways to choose 3 white balls} = \binom{4}{3} = 4. \]

2. **All red:**
\[ \text{Ways to choose 3 red balls} = \binom{5}{3} = 10. \]

3. **All black:**
\[ \text{Ways to choose 3 black balls} = \binom{6}{3} = 20. \]

Adding these cases together gives:

\[ \text{Total ways for all same color} = 4 + 10 + 20 = 34. \]

Now, the probability that all balls drawn are of the same color is:

\[ P(\text{all same color}) = \frac{34}{455}. \]

### Summary of Results

1. Probability that no ball drawn is black: \( \frac{84}{455} \).
2. Probability that exactly 2 are black: \( \frac{135}{455} \).
3. Probability that all are of the same color: \( \frac{34}{455} \).