Let [tex]H[/tex] be the number of times a coin lands heads up. Then the coin lands tails up [tex]9-H[/tex] times, and [tex]X=H-(9-H)=2H-9[/tex]. [tex]H[/tex] follows a binomial distribution with [tex]p=0.5[/tex] and [tex]n=9[/tex], so that
[tex]P(H=h)=\begin{cases}\dbinom9h0.5^9&\text{for }h\in\{0,1,2,\ldots,9\}\\0&\text{otherwise}\end{cases}[/tex]
Then we have
[tex]P(X=0)=P(2H-9=0)=P(H=4.5)=0[/tex]
because [tex]H[/tex] can only take on integer values. The other probability is
[tex]P(X=3)=P(2H-9=3)=P(H=6)\approx0.1641[/tex]
In terms of [tex]H[/tex], we have [tex]Y=2H-8[/tex] and [tex]H[/tex] follows a binomial distribution with [tex]n=8[/tex] and the same probability [tex]p[/tex] as before, so that
[tex]P(H=h)=\begin{cases}\dbinom8h0.5^8&\text{for }h\in\{0,1,2,\ldots,8\}\\0&\text{otherwise}\end{cases}[/tex]
Then we find
[tex]P(Y=0)=P(2H-8=0)=P(H=4)\approx0.2734[/tex]
and
[tex]P(Y=3)=P(2H-8=3)=P(H=5.5)=0[/tex]
