Probabilities in a binomial distribution not summing to one? I am having an issue with what seems to me to be a quite simple binomial distribution problem:
There are five members in my family, two of whom are male. My question is: what are the probabilities of choosing 0, 1 and 2 males if the number of trials ($n$) is three?
My assumptions are as follows:


*

*Probability for success ($p$) is $0.4$ (two males divided by five total family members)

*The compliment for that probability is $0.6$ ($1 - p$)

*The number of trials per experiment ($n$) is $3$ (i.e. I am picking three members of the family at random each time)


The formula I am using for the binomial distribution is this:
$$P(x) = \frac{N!}{x!  (N-x)!}  p ^ x q^{(N-x)}$$
The issue I am coming across is that when I calculate the probabilities of the outcomes (0, 1, 2) I receive the following outputs respectively:


*

*$0.2160$

*$0.4320$

*$0.2880$


These outputs sum to only $0.9360$. However, when I include a third $x$ ($x = 3$), I receive a probability of $0.0640$. when this probability is included the sum of all probabilities is 1.
My question is how can this be possible in a situation where there are only 2 males in the family of 5? Shouldn't the probability of picking 3 males be impossible (i.e. zero probability)?
Am I misunderstanding the nature of my problem? Or the nature of the binomial distribution?
 A: The binomial distribution models a repeated experiment with two outcomes: success and failure. The experiment has to be identical each time you repeat it. The classic example is a coin flip: each time you flip the coin, you can either get heads or tails, and the two are equally likely. Flipping the coin the 5th time is identical to flipping it the 1st time.
In this case, if the experiment is to be repeated identically, you need to start with all 5 family members each time. That means you might end up choosing the same person twice -- or thrice. So 3 males is a possibility.
If you're trying to model choosing one family member from 5, and then choosing another from the remaining 4, and then another from the remaining 3, the experiment isn't identically repeated each time: the first time you choose one from 5, and the second time you choose one from 4. So the binomial distribution won't work.
A: You're misunderstanding binomial distribution. In this kind of distribution (binomial) the probability remains constant. And that's not your case. When you pick a male in the first turn, the probability of picking another male in the second turn is $0.5$ ($2$ male and $2$ female left). And so on for the different cases and turns.
So, your problem is not about binomial distribution.
A: The binomial distribution assumes that you pick the objects (in this case, family members) with replacement. In a sample of 3 it is entirely possible that a male is picked three times, since each time the picked male is returned to the population. Indeed, your calculated probability of 0.0640 for this scenario is correct.
If you want the probability of picking three males to be zero, you have to pick the sample without replacement. In this case the distribution changes to hypergeometric, where the probability of getting $k$ successes in a sample of $n$ from a population with $K$ successes in $N$ is $\frac{\binom Kk\binom{N-K}{n-k}}{\binom Nn}$. For this model, the probabilities of picking 0, 1 and 2 males become $\frac1{10}$, $\frac35$ and $\frac3{10}$ respectively.
