You are making cookies and add N chips to dough randomly, and split it into 100 equal cookies, again at random. How many chips should go into dough? Question: You are making chocolate chip cookies. You add N chips randomly to the dough and you randomly split the dough into 100 equal cookies. How many chips should go into the dough to give a probability of at least 90% that every cookie has at least one chip?
I tried to attempt to solve this using IID random variables. I am not sure how to set the problem up. I know that there should at least be 100 chocolate chips or else the cookies will not meet the "at least 1 chip per cookie" requirement and that there is 10% chance that the cookies do not have a chip.
 A: There seem to be several interpretations of what you are looking for.
I have revised my answer to make it clear what I am answering, and appended a comment for a different interpretation.
I'd suggest an approximate Poisson model.  If there are $N$ chips in the dough,
then the number of chips in a random cookie is $X \sim Pois(\lambda = N/100).$
You want a 90% probability that a random cookie has at least
one chip, then solve for $N$ in $$1 - P\{X = 0\} = 1 - e^{-N/100} = .90.$$
Addendum: If you want a 90% chance that every one of the 100 cookies has
at least one chip, then that's another problem, to which the
Coupon Collection approach suggested in another Answer is a specific idea that may be
appropriate. There are Comments with
strong opinions about approaches, but no specific suggestions yet.
A: Hint: Look up the "coupon collector's" problem and use the probability estimates derived for that problem.
A: The general approach is as follows. This is only an approximation but it's an okay one as we see at the end.
Let's define $P_c$ as the probability that every cookie has at least 1 chip. Then we have the complement of this $P_{nc} = 1-P_c$ which is the probability that one or more cookies don't have a chip. We can now represent this event as a sum of simple events. Define $C_i$ as the event that the i-th cookie has no chips. We can now write
$$\begin{align}
P_{nc} & =  P(C_0 \cup C_1 \cup \cdots \cup C_{100}) \\ 
& = P \left( \bigcup_{i=1}^{100} C_i \right) \leq \sum_{i=1}^{100} P(C_i)
\end{align}$$
and we use Boole's inequality to place an upper bound on this. So now what's left is to determine $P(C_i)$ probability of a single cookie of having no chips. Here's one way to think about it:

*

*each chip arrives independently and there is a $\frac{1}{100}$ chance it will land in this cookie $C_i$

*there is a $\frac{99}{100}$ chance it will not land in cookie $C_i$

*for a total of N chips this becomes $\frac{99}{100}^N$
Thus $P(C_i) = \frac{99}{100}^N$ is the probability of a single cookie having no chips when N chips are randomly distributed amongst 100 cookies. Continuing, we have
$$\begin{align}
P_{nc} & =  \sum_{i=1}^{100} P(C_i) \\
& = \sum_{i=1}^{100} \frac{99}{100}^N = 100 \frac{99}{100}^N
\end{align}$$
Consequently $P_c \ge 1 - 100 \frac{99}{100}^N$, which we solve by taking the equality and obtain
$$\begin{align}
1 - 100 \frac{99}{100}^N & = 0.9 \\
\frac{99}{100}^N &=  1000^{-1} \\ 
N \ln{\frac{99}{100}}  & = -\ln{1000} \\ 
N & = -\frac{\ln{1000}}{\ln{\frac{99}{100}}} \\
N & \approx 687.31
\end{align}$$
Which is the upper bound solution to the problem.
Note: that by using Boole's Inequality we are assuming $C_i$ events do not overlap, i.e. we will not have more than 1 chipless cookie. These higher-order terms include probabilities of more than 1 cookie having no chips. These terms range from i=1 (1 chipless cookie, which we considered here) to 99, which represents a situation where all the chips would be in a the 100th cookie, and the remainder 99 would be chipless. It can be shown that such higher-order terms contribute little to the sum above, but I am not mathematical enough to do it.
However, numerical simulations yield a value of $N = 682-683$, so we can see it's a great approximation.
A: If there is a requirement that the chocolate chips are evenly (or "equally") distributed, then you need at least 100 chips in order for there to be a non-zero probability of having at least 1 chip in every cookie. But then there will be a 100% probability that every cookie will have a chip in it because the chips are evenly distributed, so $N=100$.
On the other hand, if you believe that the chips were "randomly" distributed, you would need to know the maximum amount of chips which could fit into 1 cookie, and assume that every cookie is the same size. You would also need a function to describe the "randomness" - i.e. if some cookies will get more chips than others and by how much.
