Question

Let X equal to the number of heads after 4 flips of a fair coin? Derive the probability mass function for X, and plot it. Also, compute the E[X] of X.

Answer 1

The probability of 0 heads is equal to the probability of 4 heads.

Since the coin is fair, the probability of no heads in 4 tosses is

$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16}$

Same is the probability of getting 4 heads.

The probability of getting 1, 2, or 3 heads is different. For example, there are four ways to get 1 head:

HTTT, THTT, TTHT, TTTH

Each of these cases has probability 1/16 as before, but there are 4 of them so the total probability of getting 1 head is 4/16.

It is a similar procedure to count the number of ways to get 2 or 3 heads. More generally, the number of ways to get x heads out of n coin tosses is given by the binomial coefficient:

$\operatorname{nCr}\left(n,x\right)$

To generalize even further, this coin tossing experiment follows the binomial distribution. If the probability of tossing a head is p then the PMF is given by

$P(X = x) = {n \choose x} p^{x}(1-p)^{n-x}$

Since p=1/2

We have

$P(X = x) = {4 \choose x} 0.5^{x}(1-0.5)^{4-x}$

$P(X = 0) = {4 \choose 0} 0.5^{0}(1-0.5)^{4-0}=1/16$

$P(X = 1) = {4 \choose 1} 0.5^{1}(1-0.5)^{4-1}=4/16$

$P(X = 2) = {4 \choose 2} 0.5^{2}(1-0.5)^{4-2}=6/16$

$P(X = 3) = {4 \choose 3} 0.5^{3}(1-0.5)^{4-3}=4/16$

$P(X = 4) = {4 \choose 4} 0.5^{4}(1-0.5)^{4-4}=1/16$

Expectation of X

$E(X)=Sum(x*P(x))=2$