PMF of X~Binomial Coin Toss

Question

Doing some practice problems from "Introduction to Probability" (Blitstein)

There are two coins, one with probability p1 of Heads and the other with probability p2 of Heads. One of the coins is randomly chosen (with equal probabilities for the two coins). It is then flipped n>=2 times. Let X be the number of times it lands Heads.

(a) Find the PMF of X.

C1 = Event that coin 1 is used (P(Heads) = p1)

C2 = Event that coin 2 is used (P(Heads) = p1)

P(X=x) = P(X=x|C1)*P(C1)+P(X=x|C2)*P(C2)

Question: Is the notation fine? You can write P(X=x|C1)? They're both events, so I'd think so?

$ = {{n}\choose{x}}p_1^x(1-p_1)^{x-n}\frac{1}{2} + {{n}\choose{x}}p_2^x(1-p_2)^{x-n}\frac{1}{2} $

Question: is the work correct?

(b) What is the distribution of X if p1 = p2?

You can add the two terms if the probabilities are the same:

$ = {{n}\choose{x}}p_1^x(1-p_1)^{x-n} $

Question: is the logic correct?

(c) Give an intuitive explanation of why X is not Binomial for p1 != p2 (its distribution is called a mixture of two Binomials). You can assume that n is large for your explanation, so that the frequentist interpretation of probability can be applied.

I am unsure about this one.

If part A is correct, I can rewrite as

$ = {n \choose x}(p_1^x(1-p_1)^(x-n)\frac{1}{2} + p_2^x(1-p_2)^(x-n)\frac{1}{2}) $

which only has 1 binomial term... so either part A is incorrect or I am not completely understanding binomial definition.

Answer 1

What makes a PMF binomial is not that there is a binomial coefficient in the formula for the probability of each value $x.$ It is not even that there is only one binomial coefficient in the formula. What makes a PMF a binomial PMF is that the probability $P(X=x)$ can be written exactly like this:

$$ \binom nx p^x (1-p)^{n-x}. $$

If you are forced to put anything else in the formula, it's not a binomial PMF.

Can $P(X=x)$ be written this way when $p_1 \neq p_2$?

---

It seems "obvious" that the answer is "no," but I have some doubts that this is sufficiently "intuitive" for the question. The suggestion to consider large values of $n$ seems to be a hint about what kind of intuition is expected. Something to consider is what the distribution of $X$ would look like if $n$ is very large.

If you have looked at the shapes of a simple binomial distribution with parameter $p$, as $n$ gets larger you get a larger chance that the value will be within a few percent of $np.$ But what you have is a mixed distribution with half a large probability of being near $np_1$ and half a large probability of being near $np_2.$ If $n$ is really large, what do you think happens to the probability that the random value will be midway between $np_1$ and $np_2,$ compared to the probability of being at $np_1$ or $np_2$?