Question

In Problem $4.5,$for $n=3,$if the coin is assumed fair, what are the probabilities associated with the values that X can take on?

Short answer

For $n=3,$

$$\begin{gathered} X=3-2t \\ X=3\left(H,H,H\right)P=\frac{1}{8} \\ X=1\left(H,T,H\right)\left(T,H,H\right)\left(H,H,T\right)P=\frac{3}{8} \\ X=-1\left(T,T,H\right)\left(T,H,T\right)\left(H,T,T\right)P=\frac{3}{8} \\ X=-3\left(T,T,T\right)P=\frac{1}{8} \end{gathered}$$

Step-by-step solution

Step1: Given Information

X - It is the difference between the number of heads and the number of tails.

Step2: Explanation

Possible outcomes of X, when a coin is tossed n times are:

If the number of tails is t, then the number of heads(h) is n-t.

$h=n-t$

Therefore X can be written as:

$X=h-t=n-t-t=n-2t$

Therefore possible outcomes of X are:

$\{n-2t\mid t\in \{0,1,\dots ,n\left\}\right\}$

Here localid="1646472975417" $n=3,$

The result is $\{3-2t\mid t\in \{0,1,2,3\left\}\right\}$

$$\begin{gathered} X=3-2t \\ X=3\left(H,H,H\right)P=\frac{1}{8} \\ X=1\left(H,T,H\right)\left(T,H,H\right)\left(H,H,T\right)P=\frac{3}{8} \\ X=-1\left(T,T,H\right)\left(T,H,T\right)\left(H,T,T\right)P=\frac{3}{8} \\ X=-3\left(T,T,T\right)P=\frac{1}{8} \end{gathered}$$

Step3: Final Result

$$\begin{gathered} X=3-2t \\ X=3\left(H,H,H\right)P=\frac{1}{8} \\ X=1\left(H,T,H\right)\left(T,H,H\right)\left(H,H,T\right)P=\frac{3}{8} \\ X=-1\left(T,T,H\right)\left(T,H,T\right)\left(H,T,T\right)P=\frac{3}{8} \\ X=-3\left(T,T,T\right)P=\frac{1}{8} \end{gathered}$$