A machine requires all five of its micro controllers to operate correctly in order to pass acceptability tests. The probability of the installed micro controller type operating correctly is 0.99.
(a) What is the probability that the machine passes acceptability tests?
My answer is as follows
P (Passes acceptability test micro controllers)
As there are 5 micro controllers P (5)
The formula we were taught to use
5C5 * $.99^5$ * (1-.99)$(5-5)$
= 1* .95
= .95
Your formula has $(5-5)$ in it, which means that unless I'm misinterpreting it, you have: $$^5C_5(0.99)^5(1-0.99)(5-5)=^5C_5(0.99)^5(1-0.99)(0)=0,$$ which is already wrong. Since all five of your controllers need to work, all you need to do is $(0.99)^5\approx0.95$ (reading it more carefully, I realize that you might be trying to say $(1-0.99)^{(5-5)}$, which is correct). In general, if $X\sim B(n,p)$, we have $$P(X=x)=~^nC_xp^{x}(1-p)^{n-x}.$$