Question

The probability of a component being defective is 0.01. There are 100 such components in a machine. Then the probability of two or more defective components in the machine is ...........

• A. \( 1 - e^{-1} \)
• B. \( 2e^{-1} \)
• C. \( 1 - 2e^{-1} \)
• D. \( e^{-1} \)

Hint:

Use the Poisson approximation to the binomial distribution when \( n \) is large and \( p \) is small. This simplifies computations significantly.

Answer 1

The Correct Option is C

Let \( X \) be the number of defective components out of 100, with the probability of a defect being \( p = 0.01 \).
Then, \( X \sim \text{Binomial}(n = 100, p = 0.01) \).
Since \( n \) is large and \( p \) is small, we can approximate this using a Poisson distribution with parameter \( \lambda = np = 1 \).
So, \( X \sim \text{Poisson}(\lambda = 1) \)
We need \( P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \)
\[ P(X = 0) = \frac{e^{-1} \cdot 1^0}{0!} = e^{-1}, P(X = 1) = \frac{e^{-1} \cdot 1^1}{1!} = e^{-1} \]
\[ P(X \geq 2) = 1 - e^{-1} - e^{-1} = 1 - 2e^{-1} \]