In the items produced by a factory there are 10% defective items A sample of 6 items is randomly chosen . Find the probability that this sample contains (i) exactly 2 defective items (ii) not more than 2 defective items (iii) at least 3 defective items .

Answer

To solve the problem, we need to use the binomial probability formula, which is given by:

\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]

where:
- \( n \) is the total number of trials (in this case, the number of items sampled),
- \( r \) is the number of successful trials (defective items),
- \( p \) is the probability of success (probability of an item being defective),
- \( q \) is the probability of failure (probability of an item being non-defective).

Given:
- The probability of a defective item \( p = 0.1 \) (10% defective),
- The probability of a non-defective item \( q = 1 - p = 0.9 \),
- The sample size \( n = 6 \).

### Step 1: Calculate the probability of exactly 2 defective items

We need to find \( P(X = 2) \):

\[ P(X = 2) = \binom{6}{2} (0.1)^2 (0.9)^{6-2} \]

Calculating \( \binom{6}{2} \):

\[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]

Now substituting back into the formula:

\[ P(X = 2) = 15 \times (0.1)^2 \times (0.9)^4 \]

Calculating \( (0.1)^2 = 0.01 \) and \( (0.9)^4 = 0.6561 \):

\[ P(X = 2) = 15 \times 0.01 \times 0.6561 = 15 \times 0.006561 = 0.098415 \]

Thus, the probability of exactly 2 defective items is approximately **0.0984**.

### Step 2: Calculate the probability of not more than 2 defective items

We need to find \( P(X \leq 2) \):

\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]

Calculating \( P(X = 0) \):

\[ P(X = 0) = \binom{6}{0} (0.1)^0 (0.9)^6 = 1 \times 1 \times 0.531441 = 0.531441 \]

Calculating \( P(X = 1) \):

\[ P(X = 1) = \binom{6}{1} (0.1)^1 (0.9)^5 = 6 \times 0.1 \times 0.59049 = 6 \times 0.059049 = 0.354294 \]

Now, we can sum these probabilities:

\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X \leq 2) = 0.531441 + 0.354294 + 0.098415 = 0.98415 \]

Thus, the probability of not more than 2 defective items is approximately **0.9842**.

### Step 3: Calculate the probability of at least 3 defective items

We need to find \( P(X \geq 3) \):

\[ P(X \geq 3) = 1 - P(X < 3) = 1 - P(X \leq 2) \]

From Step 2, we already calculated \( P(X \leq 2) \):

\[ P(X \geq 3) = 1 - 0.98415 = 0.01585 \]

Thus, the probability of at least 3 defective items is approximately **0.0159**.

### Summary of Results:
1. Probability of exactly 2 defective items: **0.0984**
2. Probability of not more than 2 defective items: **0.9842**
3. Probability of at least 3 defective items: **0.0159**