Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired, whereas the other 40% must be replaced with new units. If a company purchases 10 of these telephones, what is the probability that exactly two will end up being replaced under warranty?
I know the formula and believe I can answer the question, I just want to make sure I'm using the right probability,p from the equation bellow.
Using the equation: $p(x)=\binom{n}{x}p^x(1-p)^{n-x}$. I wanted to know if $p=2/10$?
Best Answer
The problem states that $20\%$ are likely to need warranty service, and of the $20\%$ that need repair, $40\%$ are likely to need to be replaced. This implies that $8\%$ of the total are likely to need warranty service and be replaced $(.20)(.40) = 0.08$. Therefore, $p=0.08$, $n = 10$, and $x=2$. We now plug all of this into formula $$ b(x;n,p) = {n\choose x}p^x(1-p)^{n-x} $$ and we get $$ b(2;10,0.08)={10\choose2}(0.08)^2(1-0.08)^{10-2}=45(0.0064)(0.92)^8\approx 0.14781. $$