Four rotten oranges are accidentally mixed with 16 good ones. Three oranges are drawn at random from the mixed lot. Let X be the number of rotten oranges drawn. Find the mean and variance of X.

Answer

To solve the problem of finding the mean and variance of the random variable \( X \), which represents the number of rotten oranges drawn from a mixed lot of 4 rotten and 16 good oranges, we will follow these steps:

### Step 1: Define the Problem
We have a total of 20 oranges (4 rotten + 16 good). We will draw 3 oranges at random and let \( X \) be the number of rotten oranges drawn. The possible values of \( X \) are 0, 1, 2, and 3.

### Step 2: Calculate the Total Number of Ways to Choose 3 Oranges
The total number of ways to choose 3 oranges from 20 is given by the combination formula:
\[ \text{Total ways} = \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140 \]

### Step 3: Calculate the Probability for Each Value of \( X \)

1. **For \( X = 0 \)** (0 rotten oranges):
- Choose 0 rotten from 4 and 3 good from 16:
\[ P(X=0) = \frac{\binom{4}{0} \cdot \binom{16}{3}}{\binom{20}{3}} = \frac{1 \cdot 560}{1140} = \frac{560}{1140} = \frac{28}{57} \]

2. **For \( X = 1 \)** (1 rotten orange):
- Choose 1 rotten from 4 and 2 good from 16:
\[ P(X=1) = \frac{\binom{4}{1} \cdot \binom{16}{2}}{\binom{20}{3}} = \frac{4 \cdot 120}{1140} = \frac{480}{1140} = \frac{8}{19} \]

3. **For \( X = 2 \)** (2 rotten oranges):
- Choose 2 rotten from 4 and 1 good from 16:
\[ P(X=2) = \frac{\binom{4}{2} \cdot \binom{16}{1}}{\binom{20}{3}} = \frac{6 \cdot 16}{1140} = \frac{96}{1140} = \frac{8}{95} \]

4. **For \( X = 3 \)** (3 rotten oranges):
- Choose 3 rotten from 4:
\[ P(X=3) = \frac{\binom{4}{3} \cdot \binom{16}{0}}{\binom{20}{3}} = \frac{4 \cdot 1}{1140} = \frac{4}{1140} = \frac{1}{285} \]

### Step 4: Create a Probability Distribution Table
| \( X \) | Probability \( P(X) \) |
|---------|----------------------------------|
| 0 | \( \frac{28}{57} \) |
| 1 | \( \frac{8}{19} \) |
| 2 | \( \frac{8}{95} \) |
| 3 | \( \frac{1}{285} \) |

### Step 5: Calculate the Mean \( E(X) \)
The mean is calculated using the formula:
\[ E(X) = \sum (X \cdot P(X)) \]
Calculating each term:
\[ E(X) = 0 \cdot \frac{28}{57} + 1 \cdot \frac{8}{19} + 2 \cdot \frac{8}{95} + 3 \cdot \frac{1}{285} \]
Calculating each term:
- \( 0 \cdot \frac{28}{57} = 0 \)
- \( 1 \cdot \frac{8}{19} = \frac{8}{19} \)
- \( 2 \cdot \frac{8}{95} = \frac{16}{95} \)
- \( 3 \cdot \frac{1}{285} = \frac{3}{285} \)

Finding a common denominator (285):
\[ E(X) = 0 + \frac{8 \cdot 15}{285} + \frac{16 \cdot 3}{285} + \frac{3}{285} = \frac{120 + 48 + 3}{285} = \frac{171}{285} = \frac{3}{5} \]

### Step 6: Calculate the Variance \( Var(X) \)
Variance is calculated using:
\[ Var(X) = E(X^2) - (E(X))^2 \]
First, we need \( E(X^2) \):
\[ E(X^2) = \sum (X^2 \cdot P(X)) \]
Calculating each term:
\[ E(X^2) = 0^2 \cdot \frac{28}{57} + 1^2 \cdot \frac{8}{19} + 2^2 \cdot \frac{8}{95} + 3^2 \cdot \frac{1}{285} \]
Calculating each term:
- \( 0^2 \cdot \frac{28}{57} = 0 \)
- \( 1^2 \cdot \frac{8}{19} = \frac{8}{19} \)
- \( 2^2 \cdot \frac{8}{95} = \frac{32}{95} \)
- \( 3^2 \cdot \frac{1}{285} = \frac{9}{285} \)

Finding a common denominator (285):
\[ E(X^2) = 0 + \frac{8 \cdot 15}{285} + \frac{32 \cdot 3}{285} + \frac{9}{285} = \frac{120 + 96 + 9}{285} = \frac{225}{285} = \frac{15}{19} \]

Now substituting into the variance formula:
\[ Var(X) = E(X^2) - (E(X))^2 = \frac{15}{19} - \left(\frac{3}{5}\right)^2 = \frac{15}{19} - \frac{9}{25} \]
Finding a common denominator (475):
\[ Var(X) = \frac{15 \cdot 25}{475} - \frac{9 \cdot 19}{475} = \frac{375 - 171}{475} = \frac{204}{475} \]

### Final Answers
- Mean \( E(X) = \frac{3}{5} \)
- Variance \( Var(X) = \frac{204}{475} \)