It is know that a box of 8 batteries contains 3 defective pieces and a person randomly selects 2 batteries form this box. Find the probability distrubtion of the number of defective batteries.

Answer

To solve the problem of finding the probability distribution of the number of defective batteries when selecting 2 batteries from a box of 8 (which contains 3 defective and 5 non-defective), we will follow these steps:

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

### Step 2: Calculate Total Outcomes
The total number of ways to choose 2 batteries from 8 is given by the combination formula:
\[ \text{Total outcomes} = \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]

### Step 3: Calculate Probability for Each Case
We will calculate the probability for each possible value of \( X \).

#### Case 1: \( X = 0 \) (0 defective batteries)
To select 0 defective batteries, we select 2 from the 5 non-defective batteries:
\[ P(X = 0) = \frac{\binom{3}{0} \cdot \binom{5}{2}}{\binom{8}{2}} = \frac{1 \cdot 10}{28} = \frac{10}{28} = \frac{5}{14} \]

#### Case 2: \( X = 1 \) (1 defective battery)
To select 1 defective battery, we select 1 from the 3 defective and 1 from the 5 non-defective:
\[ P(X = 1) = \frac{\binom{3}{1} \cdot \binom{5}{1}}{\binom{8}{2}} = \frac{3 \cdot 5}{28} = \frac{15}{28} \]

#### Case 3: \( X = 2 \) (2 defective batteries)
To select 2 defective batteries, we select both from the 3 defective:
\[ P(X = 2) = \frac{\binom{3}{2} \cdot \binom{5}{0}}{\binom{8}{2}} = \frac{3 \cdot 1}{28} = \frac{3}{28} \]

### Step 4: Summarize the Probability Distribution
Now we can summarize the probability distribution of \( X \):
- \( P(X = 0) = \frac{5}{14} \)
- \( P(X = 1) = \frac{15}{28} \)
- \( P(X = 2) = \frac{3}{28} \)

### Step 5: Verify the Probabilities Sum to 1
To ensure the probabilities are valid, we check if they sum to 1:
\[ P(X = 0) + P(X = 1) + P(X = 2) = \frac{5}{14} + \frac{15}{28} + \frac{3}{28} \]
Converting \( \frac{5}{14} \) to a fraction with a denominator of 28:
\[ \frac{5}{14} = \frac{10}{28} \]
Now summing:
\[ \frac{10}{28} + \frac{15}{28} + \frac{3}{28} = \frac{28}{28} = 1 \]

### Final Probability Distribution Table
The final probability distribution can be represented as follows:

| Number of Defective Batteries (X) | Probability \( P(X) \) |
|-------------------------------------|--------------------------|
| 0 | \( \frac{5}{14} \) |
| 1 | \( \frac{15}{28} \) |
| 2 | \( \frac{3}{28} \) |