The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.05 flaw per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.
(a) What is the probability that there are no surface flaws in an auto’s interior?
(b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws?
(c) If 10 cars are sold to a rental company, what is the probability that at most one car has any surface flaws?
Formulas
Poisson
$$P(X=x) = P\left( x \right) = \frac{{e^{ - \lambda } \lambda ^x }}{{x!}} $$
Binomial
$$\left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right)p^k (1-p)^{n - k}$$
My attempt
Let X - Random variable which counts the number of surface flaws in a car
(a) $ \lambda $ = $ 0.05 * 10 $ = 0.5
$P\left( X =0\right) = \frac{{e^{ - 0.5 } * 0.5 ^0 }}{{0!}}$ = 0.60653
I had a doubt in the part (b) of the problem
(b) It will obey a binomial distribution
Let Y = Random variable which counts the number of cars which have surface flaws
Probability that the car has surface flaws (p) = 1 - 0.60653 = 0.39347
$P(Y = 0) = \binom{10}{0} * (0.39347)^{0} * (0.60653)^{10} = (0.60653)^{10} $
Now if I try to use an alternative method , i get a different answer
Let W = Random variable which counts the number of cars which don't have surface flaws
Probability that a car has no surface flaws $ p = 0.60653 $
P(W = 10) = $\binom{10}{10} * (0.39347)^{10} * (0.60653)^{0} = (0.39347)^{10} $
Where am I going wrong?