Chapter 2 Section 2 Question 2
Question
Suppose we flip three fair coins, and let X be the number of heads showing.
  1. Compute $P(X = x)$ for every real number x.
  2. Write a formula for $P(X \in \textbf{B})$, for any subset $\textbf{B}$ of the real numbers.
Answer
Author: Mohammad-Ali Bandzar| Date:Oct 14 2020
(a) Compute $P(X = x)$ for every real number x.
All possible outcomes from flipping three coins:
Heads,Heads,Heads
Heads,Heads,Tails
Heads,Tails,Heads
Heads,Tails,Tails
Tails,Heads,Heads
Tails,Heads,Tails
Tails,Tails,Heads
Tails,Tails,Tails
The probability of no heads is $\frac{1}{0}$ therefore $P(X=0)=\frac{1}{8}$
The probability of one head is $\frac{3}{8}$ therefore $P(X=1)=\frac{3}{8}$
The probability of two heads is $\frac{3}{8}$ therefore $P(X=2)=\frac{2}{8}$
The probability of three heads is $\frac{1}{8}$ therefore $P(X=3)=\frac{1}{8}$
the probability of any other number of heads is zero $P(X=x)=0$ for $x\notin \{0,1,2,3\}$
(b) Write a formula for $P(X \in \textbf{B})$, for any subset B of the real numbers.
We can simply write the probabilities we calculated above multiplied by the indicator function for that outcome, the sum of them all will represent the probability of $\textbf{B}$. $$P(X\in \textbf{B})=\frac{1}{8}\textit{I}_B(0)+\frac{3}{8}\textit{I}_B(1)+\frac{2}{8}\textit{I}_B(2)+\frac{1}{8}\textit{I}_B(3)$$ Alternatively, assuming n many coin tosses, we have $2^n$ many possible outcomes, of those outcomes we have $n \choose x$ many ideal outcomes where x represents the desired number of coins showing heads.
$$\frac{n\choose x}{2^n}$$ This equation represents the probability of having x many heads in n many coin tosses. $$P(X\in \textbf{B})=\frac{3\choose 0}{2^3}\textit{I}_B(0)+\frac{3\choose 1}{2^3}\textit{I}_B(1)+\frac{3\choose 2}{2^3}\textit{I}_B(2)+\frac{3\choose 3}{2^3}\textit{I}_B(3)$$ Note that this equation simplifies to exactly the same one as above.