Question

Defects on plywood sheet occur at random with the average of one defect per 50 sq.ft. Find the probability that such a sheet has : (a) no defect (b) at least one defect (use \(e^{-1} = 0.3678\))

Hint:

For Poisson problems, remember that "at least one" is almost always calculated as "1 minus the probability of zero." It saves you from having to calculate an infinite series of probabilities (\(P(1) + P(2) + ...\)).

Answer 1

This is an example of a Poisson distribution, which is used for modeling the number of events occurring in a fixed interval of time or space.
The average number of defects (\(\lambda\)) per 50 sq.ft. is given as 1. So, the mean of the distribution is \(m = 1\).
The Poisson probability formula is \( P(X=x) = \frac{e^{-m} m^x}{x!} \).
(a) Probability of no defect:
Here, we need to find \(P(X=0)\). \[ P(X=0) = \frac{e^{-1} (1)^0}{0!} \] Since \(1^0 = 1\) and \(0! = 1\): \[ P(X=0) = e^{-1} \] Given \(e^{-1} = 0.3678\), the probability of no defect is 0.3678.
(b) Probability of at least one defect:
"At least one defect" means \(X \geq 1\). The easiest way to calculate this is by using the complement rule: \[ P(X \geq 1) = 1 - P(X=0) \] We already calculated \(P(X=0)\) in part (a). \[ P(X \geq 1) = 1 - 0.3678 = 0.6322 \] The probability of at least one defect is 0.6322.