Question

5B Let random variable X represent the number of heads minus the number of tails when a fair coin is tossed 9 times.

Answer 1

Let $H$ be the number of times a coin lands heads up. Then the coin lands tails up $9-H$ times, and $X=H-(9-H)=2H-9$. $H$ follows a binomial distribution with $p=0.5$ and $n=9$, so that

$P(H=h)=\begin{cases}\dbinom9h0.5^9&\text{for }h\in\{0,1,2,\ldots,9\}\\0&\text{otherwise}\end{cases}$

Then we have

$P(X=0)=P(2H-9=0)=P(H=4.5)=0$

because $H$ can only take on integer values. The other probability is

$P(X=3)=P(2H-9=3)=P(H=6)\approx0.1641$

In terms of $H$, we have $Y=2H-8$ and $H$ follows a binomial distribution with $n=8$ and the same probability $p$ as before, so that

$P(H=h)=\begin{cases}\dbinom8h0.5^8&\text{for }h\in\{0,1,2,\ldots,8\}\\0&\text{otherwise}\end{cases}$

Then we find

$P(Y=0)=P(2H-8=0)=P(H=4)\approx0.2734$

and

$P(Y=3)=P(2H-8=3)=P(H=5.5)=0$