Chapter 2 Section 2 Question 2

Question

Suppose we flip three fair coins, and let X be the number of heads showing.

- Compute $P(X = x)$ for every real number x.
- 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.