If `2%` of a given lot of manufactured parts are defective , then the probability that in a sample of 100 items has no defective is

Answer

To solve the problem of finding the probability that in a sample of 100 items, there are no defective parts when 2% of the manufactured parts are defective, we can use the Poisson distribution. Here’s the step-by-step solution:

### Step 1: Determine the parameters
Given that 2% of the manufactured parts are defective, we can calculate the expected number of defective parts (λ) in a sample of 100 items.

\[ \text{Defective rate} = 2\% = \frac{2}{100} = 0.02 \]
\[ \text{Total items} = 100 \]
\[ \lambda = \text{Defective rate} \times \text{Total items} = 0.02 \times 100 = 2 \]

### Step 2: Use the Poisson distribution formula
The probability of observing \( x \) events (defective parts) in a Poisson distribution is given by the formula:

\[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \]

### Step 3: Calculate the probability of no defective parts
We need to find the probability of having no defective parts, which means \( x = 0 \).

\[ P(X = 0) = \frac{e^{-\lambda} \lambda^0}{0!} \]
Substituting \( \lambda = 2 \) and \( x = 0 \):

\[ P(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} \]

### Step 4: Simplify the expression
Since \( 2^0 = 1 \) and \( 0! = 1 \):

\[ P(X = 0) = e^{-2} \cdot 1 = e^{-2} \]

### Step 5: Final answer
Thus, the probability that in a sample of 100 items, there are no defective parts is:

\[ P(X = 0) = e^{-2} \]