Q:

Probability of Finding One of a Very Few Defective Products in a Large Population

A manufacturer knows there are $30$ defective widgets in a container of $45700$ widgets.  If he pulls $700$ without replacement, what is the probability he will get at least $1$ defective widget? At least $2$ defective widgets?

A:

For this type of question, it's almost always easier to find the difference to $1$.
If $p$ is the answer to "What is the probability he won't find any defective widgets", the first answer is $1-p$. So, let's find $p$.
The probability of finding no defective widgets with $1$ widget taken is $\frac{45670}{45700}$ since there is only $30$ defective widgets. With $2$ widgets, the probability becomes $\frac{45670}{45700}*\frac{45669}{45699}$ because when we take the second widget, one "good" widget has already be taken.
Iteratively, with 700 widgets, we get $p=\frac{45670*45699*...*44971}{45700*45699*...*45001}$ which simplifies in $p=\frac{45000*44999*...*44971}{45700*45699*...*45671}$ which is approximately $0.6293$.
The answer to the first question is $1-p$ so $0.3707 = 37.07\%$
There was an other way to get this answer, and I'm going to use it for the second answer.
So if we have $p_1$ the probability of having exactly $1$ bad widget, the answer to the second question will be $1-p-p_1$. Let's calculate $p_1$.
There are $\binom{45700}{700}$ ways to take $700$ widgets from $45700$. There are $\binom{45670}{699}$ ways to take $699$ widgets from the good $45670$. Finally, there are $\binom{30}{1}$ ways of taking $1$ bad widget from the bad $30$ ones.
As a consequence, we have $p_1 = \frac{\binom{45670}{699}\binom{30}{1}}{\binom{45700}{700}}$ which is approximately $0.2938 = 29.38\%$.
We can conclude : the answer to the second question is $1-p-p_1 = 7.69\%$

