Four bad eggs are accidentally mixed with 10 good ones. Three eggs are drawn simultaneously at random. Find the mean and variance of the bad eggs drawn.

Answer

To solve the problem of finding the mean and variance of the number of bad eggs drawn when four bad eggs are mixed with ten good eggs, we can follow these steps:

### Step 1: Define the Random Variable
Let \( X \) be the random variable representing the number of bad eggs drawn. The possible values of \( X \) are \( 0, 1, 2, \) and \( 3 \).

### Step 2: Total Number of Eggs
The total number of eggs is:
\[ \text{Total eggs} = \text{Bad eggs} + \text{Good eggs} = 4 + 10 = 14 \]

### Step 3: Calculate Probabilities
We need to calculate the probabilities for each value of \( X \).

#### Probability of Drawing 0 Bad Eggs (\( P(X=0) \))
To draw 0 bad eggs, we must draw all good eggs:
\[ P(X=0) = \frac{\binom{10}{3}}{\binom{14}{3}} = \frac{120}{364} = \frac{30}{91} \]

#### Probability of Drawing 1 Bad Egg (\( P(X=1) \))
To draw 1 bad egg, we choose 1 from the 4 bad eggs and 2 from the 10 good eggs:
\[ P(X=1) = \frac{\binom{4}{1} \cdot \binom{10}{2}}{\binom{14}{3}} = \frac{4 \cdot 45}{364} = \frac{180}{364} = \frac{45}{91} \]

#### Probability of Drawing 2 Bad Eggs (\( P(X=2) \))
To draw 2 bad eggs, we choose 2 from the 4 bad eggs and 1 from the 10 good eggs:
\[ P(X=2) = \frac{\binom{4}{2} \cdot \binom{10}{1}}{\binom{14}{3}} = \frac{6 \cdot 10}{364} = \frac{60}{364} = \frac{15}{91} \]

#### Probability of Drawing 3 Bad Eggs (\( P(X=3) \))
To draw 3 bad eggs, we choose all 3 from the 4 bad eggs:
\[ P(X=3) = \frac{\binom{4}{3}}{\binom{14}{3}} = \frac{4}{364} = \frac{1}{91} \]

### Step 4: Calculate Mean (\( E(X) \))
The mean \( E(X) \) is calculated as follows:
\[ E(X) = \sum_{x=0}^{3} x \cdot P(X=x) = 0 \cdot \frac{30}{91} + 1 \cdot \frac{45}{91} + 2 \cdot \frac{15}{91} + 3 \cdot \frac{1}{91} \]
\[ E(X) = 0 + \frac{45}{91} + \frac{30}{91} + \frac{3}{91} = \frac{78}{91} \]

### Step 5: Calculate Variance (\( Var(X) \))
To calculate variance, we first need \( E(X^2) \):
\[ E(X^2) = \sum_{x=0}^{3} x^2 \cdot P(X=x) = 0^2 \cdot \frac{30}{91} + 1^2 \cdot \frac{45}{91} + 2^2 \cdot \frac{15}{91} + 3^2 \cdot \frac{1}{91} \]
\[ E(X^2) = 0 + \frac{45}{91} + \frac{60}{91} + \frac{9}{91} = \frac{114}{91} \]
Now, we can find the variance:
\[ Var(X) = E(X^2) - (E(X))^2 = \frac{114}{91} - \left(\frac{78}{91}\right)^2 \]
Calculating \( \left(\frac{78}{91}\right)^2 \):
\[ \left(\frac{78}{91}\right)^2 = \frac{6084}{8281} \]
Now substituting back:
\[ Var(X) = \frac{114}{91} - \frac{6084}{8281} = \frac{114 \cdot 8281 - 6084 \cdot 91}{91 \cdot 8281} \]
Calculating the numerator:
\[ 114 \cdot 8281 = 944154, \quad 6084 \cdot 91 = 553644 \]
So,
\[ Var(X) = \frac{944154 - 553644}{91 \cdot 8281} = \frac{390510}{91 \cdot 8281} \]

### Final Results
The mean and variance of the number of bad eggs drawn are:
\[ \text{Mean} = \frac{78}{91}, \quad \text{Variance} = \frac{390510}{91 \cdot 8281} \]