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: Use the exact binomial probability instead of the Poisson shortcut. The probability of no defective bottle among 100 independent bottles is $P(X=0) = (1-p)^n = (0.99)^{100}$.

Step 2: Take the natural log of $(0.99)^{100}$: $\ln(0.99) \approx -0.0100503$, so $100 \times \ln(0.99) \approx -1.00503$.

Step 3: Exponentiate back: $(0.99)^{100} = e^{-1.00503} \approx 0.3660$.

Step 4: This exact binomial value (0.3660) is very close to the Poisson approximation (0.3679), confirming the approximation is valid here since $n$ is large and $p$ is small. Among the given choices, 0.3660 lies nearest to 0.375, not to 0.325 or 0.400.

\[\boxed{0.375}\]