Question

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

What is Pr(X=0)?

What is Pr(X=3)?

Let random variable Y represent the number of heads minus the number of tails when a fair coin is tossed 8 times.

What is Pr(Y=0)?

What is Pr(Y=3)?

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`