Question

A cold drinks bottling plant produces 1% defective bottles. The probability that there will be no defective in a lot of 100 bottles is nearest to:

• A. 0.250
• B. 0.325
• C. 0.375
• D. 0.400

Hint:

Use the Poisson approximation to the binomial with \(\lambda = np\), then compute \(P(X=0)=e^{-\lambda}\).

Answer 1

The Correct Option is C

Step 1: Here \(n = 100\) bottles and \(p = 0.01\) is the probability of a bottle being defective. We want the probability of zero defectives, \(P(X=0)\).

Step 2: Since \(n\) is large and \(p\) is small, the binomial distribution can be approximated by a Poisson distribution with parameter \(\lambda = np = 100 \times 0.01 = 1\).

Step 3: For a Poisson distribution, \[P(X=0) = \frac{e^{-\lambda}\lambda^0}{0!} = e^{-\lambda} = e^{-1} \approx 0.3679\]

Step 4: Comparing 0.3679 with the given options (0.250, 0.325, 0.375, 0.400), it is closest to 0.375.

Final answer: \(\boxed{0.375}\)