What is the probability of 4 out of 25 are intoxicated Q: According to an accurate survey, 40% of people checked at the exit of a well-known pub have made excessive use of alcoholic drinks. If we take a random sample of 25 persons, what is the probability that at least 4 of them are flagged?
 A: We use the following model. We check, one after the other, $25$ people leaving the pub, and determine for each whether she is intoxicated. We assume that for any person, the probability of intoxication is $0.40$, and that the intoxication/sobriety of the various people are independent events. (This assumption is rather dubious.)
Under this model, the number $X$ of intoxicated people in a sample of $25$ has binomial distribution, and for any $k$, we have that the probability that $X$ is exactly $k$ is given by
$$\Pr(X=k)=\binom{25}{k}(0.4)^k (0.6)^{25-k}.\tag{1}$$
We want $\Pr(X\ge 4)$. Now use Formula (1). Unless you are using software, it will be substantially easier to calculate first the probability that $X\le 3$. 
A: Use the Binomial Distrubituion, and sum for each possible value, since the question says that 4 or more could be flagged
$(nCr)p^k(1-p)^m$ where $m=n-k$
So this would be the probability that 4 are flagged, 5 are flagged and so on. Since there are many numbers, it would be much easier to express this as $1-P(3)-P(2)-P(1)-P(0)$. I have to run, so I can't write this whole thing out, so I'll just leave an example
For $P(3)$, we would write 25$C$3$(0.4)^3$$(0.6)^{22}$   $~$$~$ 
Repeat this for each of the other numbers, and subtract from 1.
