Probability that the sum of identically distributed Bernoulli random variables equalling some number.

Question

I'm solving a problem where it seems that I am missing some information. The problem is: Suppose you randomly select 100 Bernoulli random variables, $X_i$. What is the probability that $P(Y = 20)$ where $Y = X_1 + \ldots + X_{100}$. What is the probability that $X_i = 1$?

I am confused by this question because it seems we are not given sufficient information to solve it. If we are given that $X_i$'s are IID, then $Y$ is binomially distributed. But we are not given IID, although it seems reasonable to assume IID, but we still don't know the probability that $X_i = 1$.

Is it possible to obtain a numerical solution to this problem when we don't know $P(Y = 20)$ and $P(X_i = 1)$?

Answer 1

This is a guess as far as interpretation goes. But, I don't think the following interpretation is too unlikely. Consider the following question. 100 times in a row, pick a random number $\alpha \sim \operatorname{Unif}(0, 1)$ independently of past trials, and then take a Bernoulli random variable with probability of success $\alpha$.

If that's right, the second question makes sense and is quite simple. Because, the probability $X_i = 1$ in total will simply be the average probability ranging over all $\alpha$, since our distribution is uniform. This would give us the answer to the second question of simply

$$ \int_0^1 \alpha \ \text{d} \alpha = \left.\left[\frac{1}{2}\alpha^2\right]\right|_0^1 = \boxed{\frac{1}{2}}$$

If this interpretation is correct, the first question, at least to me, is not quite as obvious at all. But, it actually is. Because each $X_i$, in this interpretation, is either 0 or 1, and it has, in total, equal odds of being either. And, since the $\alpha$'s are drawn independently, these are completely independent. So, this process is just a bernoulli trial, and the total distribution is just a binomial distribution with $p = q = .5$. In this case, it is trivial to see that the answer to the first question is simply $$ 2^{-100}{100 \choose 20} = \boxed{\frac{267991685201904841485}{633825300114114700748351602688} \approx 4.2281632676 \cdot 10^{-10}}$$

This interpretation of the ambiguity in the question is perhaps unlikely, but possible.

Answer 2

The definition of a Bernoulli random variable $X$ requires the definition of probability $P(X=0)=p$. In that case $P(X_i=1)=1-p$. As you observed binomial and assuming iid, then $P(Y=20)=\binom{100}{20}(1-p)^{20}p^{80}$.