Probability & Statistics Ch.2 Section 2 Question 2

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

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

a) Compute $P(X = x)$ for every real number x. b) Write a formula for $P(X \in B)$ , for any subset B of the real numbers.

Solution

(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 $\frac{1}{8}$ , 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 $P(X=2)=\frac{3}{8}$

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 $P(X \in B)$ , 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 B.

P(X\in B)=\frac{1}{8}\textit{I}_B(0)+\frac{3}{8}\textit{I}_B(1)+\frac{3}{8}\textit{I}_B(2)+\frac{1}{8}\textit{I}_B(3)

Alternatively, assuming n coin tosses, we have $2^n$ possible outcomes. Of those outcomes, we have $\binom{n}{x}$ ideal outcomes where x represents the desired number of coins showing heads:

\frac{\binom{n}{x}}{2^n}

This equation represents the probability of having x heads in n coin tosses:

P(X\in B)=\frac{\binom{3}{0}}{2^3}\textit{I}_B(0)+\frac{\binom{3}{1}}{2^3}\textit{I}_B(1)+\frac{\binom{3}{2}}{2^3}\textit{I}_B(2)+\frac{\binom{3}{3}}{2^3}\textit{I}_B(3)

Note that this equation simplifies to exactly the same one as above.