This problem can be modeled using a Poisson distribution since it involves counting the number of events (defective products) in a fixed interval (one day) with a known mean rate ($\lambda = 15$).

The probability mass function (PMF) for the Poisson distribution is given by:

$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$

Where:

• $\lambda = 15$ (mean number of defective products),
• $k = 14$ (the desired number of defective products),
• $e$ is the base of the natural logarithm (approximately 2.71828).

Step 1: Plug in the values

$P(X = 14) = \frac{15^{14} e^{-15}}{14!}$

Step 2: Compute the components

1. $15^{14}$
2. $14!$ (factorial of 14)
3. $e^{-15}$

Let me calculate these values.import math

Given values

lambda_value = 15 # mean k = 14 # desired number of defective products

Calculate Poisson probability

probability = (lambda_value**k * math.exp(-lambda_value)) / math.factorial(k) round(probability, 3)0.1020.102successThe probability that on any given day there are exactly 14 defective products is 0.102 (rounded to three decimal places).

Would you like a detailed breakdown of the calculation or have any further questions?

Here are 5 related questions to expand your understanding:

1. How would the probability change if the mean number of defective products was 20 instead of 15?
2. What is the probability of having fewer than 10 defective products in a day?
3. How can you calculate the cumulative probability for a range of defective product counts?
4. What are the conditions under which a Poisson distribution is appropriate?
5. How does increasing the mean ($\lambda$) affect the shape of the Poisson distribution?

Tip: The Poisson distribution assumes that the events occur independently and at a constant rate, making it perfect for modeling rare events!