A bag contains 3 red and 5 black balls. Two balls are drawn at random. The probability that one ball is red and other is black is P then the value of 28P - 15 is

Options

A. 1
B. 0
C. `1/2`
D. 2

Answer

To solve the problem step by step, we will calculate the probability of drawing one red ball and one black ball from a bag containing 3 red and 5 black balls.

### Step-by-Step Solution:

1. **Identify the Total Number of Balls:**
- The bag contains 3 red balls and 5 black balls.
- Total number of balls = 3 + 5 = 8.

2. **Determine the Total Outcomes for Drawing 2 Balls:**
- The total number of ways to choose 2 balls from 8 is given by the combination formula \( \binom{n}{r} \), which is calculated as:
\[ \text{Total Outcomes} = \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28. \]

3. **Calculate the Favorable Outcomes for Drawing 1 Red and 1 Black Ball:**
- The number of ways to choose 1 red ball from 3 is:
\[ \binom{3}{1} = 3. \]
- The number of ways to choose 1 black ball from 5 is:
\[ \binom{5}{1} = 5. \]
- Therefore, the total number of favorable outcomes for drawing 1 red and 1 black ball is:
\[ \text{Favorable Outcomes} = \binom{3}{1} \times \binom{5}{1} = 3 \times 5 = 15. \]

4. **Calculate the Probability \( P \):**
- The probability \( P \) of drawing one red ball and one black ball is given by:
\[ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{15}{28}. \]

5. **Calculate the Value of \( 28P - 15 \):**
- Now, substituting the value of \( P \):
\[ 28P - 15 = 28 \times \frac{15}{28} - 15 = 15 - 15 = 0. \]

### Final Answer:
The value of \( 28P - 15 \) is \( 0 \).