Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement.Let the random variable `X` denote the number of rotten apples.If `mu` and `alpha^(2)` represent mean and variance of `X,` respectively,then `10(mu^(2)+alpha^(2))` is equal to

Options

A. `250`
B. `20`
C. `25`
D. `30`

Answer

To solve the problem, we need to calculate the mean (μ) and variance (α²) of the random variable \(X\), which denotes the number of rotten apples drawn when four apples are drawn from a mix of three rotten and seven good apples.

### Step 1: Define the total number of apples
We have:
- Rotten apples = 3
- Good apples = 7
- Total apples = 3 + 7 = 10

### Step 2: Define the random variable \(X\)
The random variable \(X\) can take values 0, 1, 2, or 3, as we cannot draw more than 3 rotten apples (since there are only 3 rotten apples).

### Step 3: Calculate the probabilities for \(X\)
We will calculate the probabilities for each value of \(X\):

1. **For \(X = 0\)** (0 rotten apples drawn):
\[ P(X = 0) = \frac{\binom{7}{4}}{\binom{10}{4}} = \frac{35}{210} = \frac{1}{6} \]

2. **For \(X = 1\)** (1 rotten apple drawn):
\[ P(X = 1) = \frac{\binom{3}{1} \cdot \binom{7}{3}}{\binom{10}{4}} = \frac{3 \cdot 35}{210} = \frac{105}{210} = \frac{1}{2} \]

3. **For \(X = 2\)** (2 rotten apples drawn):
\[ P(X = 2) = \frac{\binom{3}{2} \cdot \binom{7}{2}}{\binom{10}{4}} = \frac{3 \cdot 21}{210} = \frac{63}{210} = \frac{3}{10} \]

4. **For \(X = 3\)** (3 rotten apples drawn):
\[ P(X = 3) = \frac{\binom{3}{3} \cdot \binom{7}{1}}{\binom{10}{4}} = \frac{1 \cdot 7}{210} = \frac{7}{210} = \frac{1}{30} \]

### Step 4: Summarize the probabilities
Now we summarize the probabilities:
- \(P(X = 0) = \frac{1}{6}\)
- \(P(X = 1) = \frac{1}{2}\)
- \(P(X = 2) = \frac{3}{10}\)
- \(P(X = 3) = \frac{1}{30}\)

### Step 5: Calculate the mean (μ)
The mean \(μ\) is calculated as:
\[ μ = \sum_{x=0}^{3} x \cdot P(X = x) \]
Calculating this:
\[ μ = 0 \cdot \frac{1}{6} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{1}{30} \]
\[ = 0 + \frac{1}{2} + \frac{6}{10} + \frac{3}{30} \]
Converting all terms to a common denominator (30):
\[ = 0 + \frac{15}{30} + \frac{18}{30} + \frac{3}{30} = \frac{36}{30} = \frac{6}{5} \]

### Step 6: Calculate the variance (α²)
The variance \(α²\) is calculated as:
\[ α² = E(X^2) - μ^2 \]
First, we need to calculate \(E(X^2)\):
\[ E(X^2) = \sum_{x=0}^{3} x^2 \cdot P(X = x) \]
Calculating this:
\[ E(X^2) = 0^2 \cdot \frac{1}{6} + 1^2 \cdot \frac{1}{2} + 2^2 \cdot \frac{3}{10} + 3^2 \cdot \frac{1}{30} \]
\[ = 0 + \frac{1}{2} + 4 \cdot \frac{3}{10} + 9 \cdot \frac{1}{30} \]
\[ = 0 + \frac{15}{30} + \frac{12}{10} + \frac{9}{30} \]
Converting \(4 \cdot \frac{3}{10}\) to a common denominator (30):
\[ = 0 + \frac{15}{30} + \frac{36}{30} + \frac{9}{30} = \frac{60}{30} = 2 \]

Now, substituting back into the variance formula:
\[ α² = E(X^2) - μ^2 = 2 - \left(\frac{6}{5}\right)^2 = 2 - \frac{36}{25} = \frac{50}{25} - \frac{36}{25} = \frac{14}{25} \]

### Step 7: Calculate \(10(μ^2 + α^2)\)
Now we calculate:
\[ 10(μ^2 + α^2) = 10\left(\left(\frac{6}{5}\right)^2 + \frac{14}{25}\right) \]
Calculating \(μ^2\):
\[ μ^2 = \left(\frac{6}{5}\right)^2 = \frac{36}{25} \]
Now adding:
\[ μ^2 + α^2 = \frac{36}{25} + \frac{14}{25} = \frac{50}{25} = 2 \]
Finally:
\[ 10(μ^2 + α^2) = 10 \cdot 2 = 20 \]

### Final Answer
Thus, the value of \(10(μ^2 + α^2)\) is **20**.