Question

A machine has a 98% probability of producing a part within acceptable tolerance levels. The machine makes 25 parts an hour. What is the probability that there are 23 or fewer acceptable parts? Round to the nearest thousandth.

Answer 1

The probability of an acceptable part is \(P_A=0.98\). The probability of a failed part is \(P_F=0.02\).
Using the binomial distribution (see - http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function) , the probability of 23 or less acceptable parts is
$$
\large
P=\sum_{n=0}^{23}{25\choose n}(P_A)^n(P_F)^{25-n}\\
=1-\sum_{n=24}^{25}{25\choose n}(P_A)^n(P_F)^{25-n}\\
={25!\over24!1!}(0.98)^{24}(0.02)+{25!\over 25!0!}(.98)^{25}\\
$$
Does this make sense?