Question

A company manufactures 4000 cars in a year with a probability that a part will be defective is 0.002. Find the probability that company produces:
a) 4 cars with defective b) at least 4 cars c) at most 4 cars with defective.

Answer 1

Answer on Question #45530, Engineering, SolidWorks

A company manufactures 4000 cars in a year with a probability that a part will be defective is 0.002. Find the probability that company produces a) 4 cars with defective b) at least 4 cars c) at most 4 cars with defective.

Solution

Here we are dealing with binomial distribution and hence we will use formula for probability mass function:

$f(k;n,p)=\Pr(X=k)={n\choose k}p^{k}(1-p)^{n-k}$

where n=4000 is number of randomly selected cars, k=4 is number of defective cars and p=0.002 is probability. a) So, the probability to find 4 defective cars among 4000 is

$Pr(X=4)={4000\choose 4}\cdot 0.002^{4}\cdot(1-0.002)^{3}996=0.0571663$

c) Here we have to add probabilities that 0 or 1 or 2 or 3 or 4 cars are defective

$P=Pr(X=0)+Pr(X=1)+Pr(X=2)+Pr(X=3)+Pr(X=4)=$

$=0.0003328+0.0026676+0.010689+0.028547+0.0571663=0.099403$

b) To find that at least cars are defective is the same that to find sum of probabilities that 4 or 5 or … 3999 or 4000 cars are defective. Or, to find 1 - (0 or 1 or 2 or 3 cars are defective). Hence

$P=1-Pr(X=0)-Pr(X=1)-Pr(X=2)-Pr(X=3)=$

$=1-0.0003328-0.0026676-0.010689-0.028547\approx 0.95776$

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