Question

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.07 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

(c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws? Round your answers to four decimal places (e.g. 98.7654).

Answer 1

For a car, lets find a probability of flaw
Here, λ = 0.07*10 = 0.7 and x = 0
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X > 0) = 1 - P(X <= 0).
P(X > 0) = 1 - (0.7^0 * e^-0.7/0!)
P(X > 0) = 1 - (0.4966)
P(X > 0) = 1 - 0.4966 = 0.5034

Now,
Here, n = 10, p = 0.5034, (1 - p) = 0.4966 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 1).
P(X <= 1) = (10C0 * 0.5034^0 * 0.4966^10) + (10C1 * 0.5034^1 * 0.4966^9)
P(X <= 1) = 0.0009 + 0.0092
P(X <= 1) = 0.0101