Q.4.6 - A First Course in Probability

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? 

Step-by-Step Solution

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 1 : Given Information

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

Step 2 : 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:

 $X=h-t=n-t-t=n-2 t$

Therefore possible outcomes of X are:

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

 

Here $n = 3,$

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

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

Step 3 : Final Result

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