A bag contains 4 white balls and 3 black balls. If two balls are drawn at random, then mean of number of white balls is

Options

A. `(8)/(7)`
B. `(7)/(8)`
C. `(1)/(4)`
D. `(1)/(5)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a bag with 4 white balls and 3 black balls, making a total of 7 balls. We need to find the mean (expected value) of the number of white balls drawn when 2 balls are drawn at random.

### Step 2: Define the Random Variable
Let \( X \) be the random variable that represents the number of white balls drawn. The possible values of \( X \) are:
- \( X = 0 \): No white balls drawn.
- \( X = 1 \): One white ball drawn.
- \( X = 2 \): Two white balls drawn.

### Step 3: Calculate the Probabilities
1. **Probability that \( X = 0 \)** (both balls are black):
- The probability of drawing a black ball first is \( \frac{3}{7} \).
- The probability of drawing a black ball second (after the first is black) is \( \frac{2}{6} \).
- Thus, the probability is:
\[ P(X = 0) = \frac{3}{7} \times \frac{2}{6} = \frac{6}{42} = \frac{1}{7} \]

2. **Probability that \( X = 1 \)** (one white and one black):
- There are two scenarios:
- First a white ball, then a black ball: \( \frac{4}{7} \times \frac{3}{6} \)
- First a black ball, then a white ball: \( \frac{3}{7} \times \frac{4}{6} \)
- Thus, the probability is:
\[ P(X = 1) = \left(\frac{4}{7} \times \frac{3}{6}\right) + \left(\frac{3}{7} \times \frac{4}{6}\right) = \frac{12}{42} + \frac{12}{42} = \frac{24}{42} = \frac{4}{7} \]

3. **Probability that \( X = 2 \)** (both balls are white):
- The probability of drawing a white ball first is \( \frac{4}{7} \).
- The probability of drawing a white ball second (after the first is white) is \( \frac{3}{6} \).
- Thus, the probability is:
\[ P(X = 2) = \frac{4}{7} \times \frac{3}{6} = \frac{12}{42} = \frac{2}{7} \]

### Step 4: Calculate the Expected Value
The expected value \( E(X) \) is calculated as follows:
\[ E(X) = \sum (X \cdot P(X)) \]
Substituting the values we found:
\[ E(X) = (0 \cdot \frac{1}{7}) + (1 \cdot \frac{4}{7}) + (2 \cdot \frac{2}{7}) \]
Calculating each term:
\[ E(X) = 0 + \frac{4}{7} + \frac{4}{7} = \frac{8}{7} \]

### Final Answer
The mean (expected value) of the number of white balls drawn is \( \frac{8}{7} \).

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