Answer to Question #336848 in Statistics and Probability for Lencho

Question

Suppose it is known that 10% of the owners of two year old automobiles have had problems with their electrical system. Compute the probability of finding exactly 2 owners that have had electrical system problems out of a group of 10 owners?

Answer 1

We have a Bernoulli trial - exactly two possible outcomes, "success" (the owner has had electrical system problems) and "failure" (the owner hasn't had electrical system problems) and the probability of success is the same every time the experiment is conducted (an owner is examined), $p=0.1, q=1-0.1=0.9, n=10.$

The probability that exactly k owners that have had electrical system problems

$P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}\cdot p^k\cdot q^{n-k}=\\ =\begin{pmatrix}10\\k\end{pmatrix}\cdot 0.1^k\cdot 0.9^{10-k}=\\ =\cfrac{10!}{k!\cdot(10-k)!}\cdot 0.1^k\cdot 0.9^{10-k}.$

The probability that exactly 2 owners that have had electrical system problems

$P(X=2)=\cfrac{10!}{2!\cdot8!}\cdot 0.1^{2}\cdot 0.9^{8}=0.1937.$