How many boxes do you need to open to find 4 spoiled oranges with a minimum of 90% probability? I'm new to probability and I got this question that I'm not sure how to go further with:
There are 50 boxes, each containing 48 Oranges, and out of all those Oranges, 72 are spoiled. The numbers given above can be considered generally, for any box of Oranges and the number of the spoiled oranges is equally distributed among the boxes. 
i) What is the probability that a box contains exactly 4 spoiled oranges?
ii) How many boxes do you need to open, in order to find a box containing exactly 4 spoiled Oranges, with a minimum probability of 90%?
So for part 1, $ 50 \times 48 = 2400$  is the number of total oranges, which means $\frac{72}{2400} = 3$% is the percentage of spoiled Oranges in general. I assume that means that an orange has a probability of 3% of being spoiled, and 97% of being good. 
Furthermore, a box has 48 Oranges, so the chance of finding a box with exactly 4 spoiled oranges is $P_n = 0.03^4 \times 0.97^{44} = 2.12 \times 10^{-7}$, is this correct, or should it be $0.03^4 + 0.97^{44}$?
For the second part, I first got the complement probability, that a box doesn't contain 4 spoiled Oranges $P_x = 1 - P_n$. My idea was that if I keep raising that to the power of n, I need to count how many times "I open the box (power of n)" until the result gets below 0.1 and then I take the total number of boxes and subtract this n. However, in this case $P_x$ is so close to one that raising it to any power of n, changes nothing. This method only worked if I used addition instead of multiplication for $P_n$ above. My $n$ was 8 in that case. 
I am not sure if my idea is the right idea. Any help would be appreciated. 
 A: For part (i), a given box contains $n = 48$ oranges.  The probability that exactly $X = 4$ of these are spoiled is a binomial probability, i.e., $$\Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x},$$ where $p = 0.03$ is the probability of a single randomly selected orange being spoiled.
For the second part, let $\theta = \Pr[X = 4]$, which you computed in the first part.  This is the chance that a single randomly selected box contains exactly $4$ spoiled oranges.  The probability that exactly $N$ boxes need to be opened in order to obtain a box with exactly $4$ spoiled oranges is a geometric probability, i.e., $$\Pr[N = n] = (1-\theta)^{n-1} \theta, \quad n = 1, 2, \ldots.$$  Thus, you wish to find the smallest positive integer $n$ such that $\Pr[N \le n] \ge 0.9$.  We note that $$\Pr[N \le n] = \sum_{k=1}^n \Pr[N = k] = \sum_{k=1}^n (1-\theta)^{k-1} \theta = \theta \frac{1 - (1-\theta)^n}{1-(1-\theta)} = 1 - (1 - \theta)^n.$$  Therefore, we can explicitly solve for $n$:  $$0.9 \le 1 - (1 - \theta)^n$$ implies $$n \ge \frac{\log 0.1}{\log (1 - \theta)},$$ for which the minimum such $n$ is $$n = \left\lceil \frac{\log 0.1}{\log (1 - \theta)} \right\rceil.$$

We should also consider the case where the oranges and boxes are finite in number as described by the problem, and that the distribution of the $72$ spoiled oranges is equiprobable among the boxes.  Then we can intuitively see that there is a nonzero probability that none of the boxes contains exactly four spoiled oranges.  What is this probability?  We can perform quick simulations to this end; on $10^6$ such simulations, I obtained the following empirical distribution of the number of boxes $Y$ with $4$ spoiled oranges: $$\begin{array}{c|c}y & \Pr[Y = y] \\ \hline
 0 & 0.099538 \\
 1 & 0.265719 \\
 2 & 0.307941 \\
 3 & 0.207353 \\
 4 & 0.088556 \\
 5 & 0.025168 \\
 6 & 0.004962 \\
 7 & 0.000691 \\
 8 & 0.000067 \\
 9 & 0.000005 \end{array}$$
Now, if we assume that we choose boxes at random without replacement, and stop at the first box containing $4$ spoiled oranges, we get the following plot of the empirical distribution of the number $N$ of boxes opened (again on $10^6$ simulations):

A: This is an exercise in the binomial distribution.  The chance of four bad oranges in one box is ${48 \choose 4}0.03^40.97^{44}$ because you choose the four oranges to be bad, then multiply the probabilities that those oranges are bad by the probabilities that the others are good.  This is a much larger chance than you calculated, which will change the answer for the second part.  You are thinking in the right direction for that one.
A: The probability that a box contains exactly four spoiled oranges, if we're assuming that a particular orange being spoiled is simply a $3\%$ probability event, would be given by $0.03^4\times 0.97^{44}\times\binom{48}{4}$. If you don't include the binomial coefficient, you're calculating the probability that the first four oranges are spoiled, and the rest are good. Since it could by any set of $4$, we need to multiply by $\binom{48}{4}$, the number of size-$4$ subsets.
That calculation comes out to about $4.126\%$, which means the second part of your calculation will work out differently, taking the complement and raising it to powers. You don't want to use addition in the second part.
