This question appears at https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1088944.html and I believe the answer to be wrong. The question is
It was reported that $9$% of the computers in a lab have not been tested for viruses. What is the probability that three randomly chosen computers were not tested?
They give the answer as:
P = (0.09)x(0.09)x(0.09) P=0.000729
The reason I believe this to be wrong is that the selection of three computers out of a fixed number of computers (at least I never attended a school with an infinite number of computers in the computer lab) is a selection without replacement and thus does not qualify for the product rule.
Further, I believe the answer to be a function of the total number of computers in the lab. Selection of $3$ out of $10$ would be much different than $3$ out of $100$ for example.
Finally, I take
three randomly chosen computers were not tested?
Should be read as none of the three computers were tested
Am I doing something wrong?
Yes, the answer is incorrect and I will prove it:
Let $a$ be the number of computers that haven't been checked for viruses and $b$ be the number of computers in total.
Then we will have:
There is a $\frac{a}{b}$ chance of choosing the first computer that haven't been checked for viruses.
There is a $\frac{a}{b-1}$ chance of choosing the second computer that haven't been checked for viruses.
There is a $\frac{a}{b-2}$ chance of choosing the third computer that haven't been checked for viruses.
The correct answer is:
$\frac{a}{b}\times\frac{a}{b-1}\times\frac{a}{b-2}=\frac{a^3}{b(b-1)(b-2)}=\frac{0.000729b}{b(b-1)(b-2)}=\frac{0.000729}{(b-1)(b-2)}$, the answer is in terms of $b$ or depends on the number of computers in total.