What is the probability of at most one flaw occurring in 225 square feet of drapery material?

To find the probability that at most one flaw occurs in 225 square feet of drapery material, we first need to determine the rate of flaws per square foot. Given that flaws appear on average once in 150 square feet, the average number of flaws (λ) in 225 square feet can be calculated as follows:

λ = (225 square feet) / (150 square feet/flaw) = 1.5 flaws

For a Poisson distribution, the probability of observing k events (in this case, flaws) is given by the formula:

P(X = k) = (e^-λ * λ^k) / k!

We want to find the probability of at most one flaw, which includes the probabilities of having 0 flaws and 1 flaw:

P(X ≤ 1) = P(X = 0) + P(X = 1)

Calculating these:

1. For 0 flaws (k = 0):

P(X = 0) = (e^-1.5 * 1.5⁰) / 0! = e^-1.5

2. For 1 flaw (k = 1):

P(X = 1) = (e^-1.5 * 1.5¹) / 1! = e^-1.5 * 1.5

Combining these:

P(X ≤ 1) = e^-1.5 + (e^-1.5 * 1.5) = e^-1.5 * (1 + 1.5)

P(X ≤ 1) = e^-1.5 * 2.5

Using the approximate value of e^-1.5 ≈ 0.2231, we can calculate:

P(X ≤ 1) ≈ 0.2231 * 2.5 ≈ 0.55775

Thus, the probability that at most one flaw occurs in 225 square feet of drapery material is approximately 0.558 or 55.8%.