Question

A certain factory turning cotter pins knows that 2% of its product is defective. If it sells cotter pins and gurantees that not more than 5 pins will be defective in a box of 100 pins. Find the approximate probability that a box will fail to meet the guranteed quality.

Answer 1

0.0166

Answer 2

The number of defective pins in a box of 100 can be modeled by a binomial distribution B(n, p), where n = 100 is the number of trials (pins) and p = 0.02 is the probability of a pin being defective.

Let X be the number of defective pins in a box. So, X ~ B(100, 0.02). The mean of this distribution is np = 100 * 0.02 = 2. Since n is large (n=100) and p is small (p=0.02), and np is moderate (np=2), the binomial distribution can be approximated by the Poisson distribution with parameter $\lambda = np = 2$. Let Y be a random variable following the Poisson distribution with $\lambda = 2$. The probability mass function of Y is $P(Y = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-2} 2^k}{k!}$.

The box fails to meet the guaranteed quality if the number of defective pins is more than 5, i.e., X > 5. Using the Poisson approximation, we want to find $P(X > 5) \approx P(Y > 5)$. $P(Y > 5) = 1 - P(Y \le 5) = 1 - [P(Y=0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5)]$.

Let's calculate the required Poisson probabilities: $P(Y=0) = \frac{e^{-2} 2^0}{0!} = e^{-2}$ $P(Y=1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2}$ $P(Y=2) = \frac{e^{-2} 2^2}{2!} = \frac{4e^{-2}}{2} = 2e^{-2}$ $P(Y=3) = \frac{e^{-2} 2^3}{3!} = \frac{8e^{-2}}{6} = \frac{4}{3}e^{-2}$ $P(Y=4) = \frac{e^{-2} 2^4}{4!} = \frac{16e^{-2}}{24} = \frac{2}{3}e^{-2}$ $P(Y=5) = \frac{e^{-2} 2^5}{5!} = \frac{32e^{-2}}{120} = \frac{4}{15}e^{-2}$

Now, sum these probabilities: $P(Y \le 5) = e^{-2} + 2e^{-2} + 2e^{-2} + \frac{4}{3}e^{-2} + \frac{2}{3}e^{-2} + \frac{4}{15}e^{-2}$ $P(Y \le 5) = e^{-2} \left(1 + 2 + 2 + \frac{4}{3} + \frac{2}{3} + \frac{4}{15}\right)$ $P(Y \le 5) = e^{-2} \left(5 + \frac{6}{3} + \frac{4}{15}\right)$ $P(Y \le 5) = e^{-2} \left(5 + 2 + \frac{4}{15}\right)$ $P(Y \le 5) = e^{-2} \left(7 + \frac{4}{15}\right)$ $P(Y \le 5) = e^{-2} \left(\frac{7 \times 15 + 4}{15}\right) = e^{-2} \left(\frac{105 + 4}{15}\right) = e^{-2} \frac{109}{15}$

To find the approximate probability, we need the value of $e^{-2}$. Using a calculator, $e^{-2} \approx 0.135335$. $P(Y \le 5) \approx 0.135335 \times \frac{109}{15} \approx 0.135335 \times 7.266667 \approx 0.983436$.

The probability that a box will fail to meet the guaranteed quality is: $P(Y > 5) = 1 - P(Y \le 5) \approx 1 - 0.983436 \approx 0.016564$.

Rounding to a few decimal places, the approximate probability is 0.0166.