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Chapter 9: Problem 561

Suppose that flaws in plywood occur at random with an average of one flaw per 50 square feet. What is the probability that a 4 foot \(\times 8\) foot sheet will have no flaws? At most one flaw? To get a solution assume that the number of flaws per unit area is Poisson distributed.

Short Answer

Expert verified
To find the probabilities of having no flaws and at most one flaw in a 4 foot 脳 8 foot plywood sheet, we first calculate the area of the sheet (32 square feet) and the average number of flaws in this area \(\lambda_{sheet} = \frac{1}{50} \times 32\). Using the Poisson probability formula, we then calculate the probability of having no flaws \(P(X=0) = \frac{\lambda_{sheet}^0 e^{-\lambda_{sheet}}}{0!}\), and the probability of having at most one flaw \(P(X \le 1) = P(X=0) + P(X=1) = P(X=0) + \frac{\lambda_{sheet}^1 e^{-\lambda_{sheet}}}{1!}\).

Step by step solution

01

Calculate the area of the plywood sheet and find the average number of flaws in that area

First, we need to find the area of the plywood sheet: Area = length 脳 width = \(4 \times 8 = \) 32 square feet Now, we can find the average number of flaws in this area: \(\lambda_{sheet} = \frac{1}{50} \times 32\)
02

Calculate the probability of having no flaws

Now that we have the average number of flaws (\(\lambda_{sheet}\)) in the given plywood sheet, we can calculate the probability of having no flaws (i.e., k = 0) using the Poisson probability formula: \(P(X=0) = \frac{\lambda_{sheet}^0 e^{-\lambda_{sheet}}}{0!}\)
03

Calculate the probability of having at most one flaw

The probability of having at most one flaw means we want to find the probability of having either 0 flaws or 1 flaw. We can calculate these probabilities separately and then add them together: - Probability of having 0 flaws (from Step 2) - Probability of having 1 flaw \(P(X=1) = \frac{\lambda_{sheet}^1 e^{-\lambda_{sheet}}}{1!}\) Adding these two probabilities together, we have: \(P(X \le 1) = P(X=0) + P(X=1)\)
04

Calculate the final probabilities

Plug in the value of \(\lambda_{sheet}\) that we calculated in Step 1 and compute the final probabilities for having no flaws and at most one flaw using the Poisson probability formula: \(P(X=0) = \frac{\lambda_{sheet}^0 e^{-\lambda_{sheet}}}{0!}\) \(P(X \le 1) = P(X=0) + P(X=1) = P(X=0) + \frac{\lambda_{sheet}^1 e^{-\lambda_{sheet}}}{1!}\) These are the desired probabilities.

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