Use Problem $9$to find $\stackrel{-}{x}$in Problem$7$.

Three coins are tossed.

The likelihood that a comparable continuous random variable has a value less than or equal to the function's argument is the value of the function.

The results in problem 9 are mentioned below.

$\begin{array}{c}{\mu }_{x-v}=E(x-y)\\ =B\left(x\right)-B\left(y\right)\end{array}$

The value of the random variable x is mentioned below.

$\begin{array}{c}{x}_1=0\\ P_1={\left(1-P\right)}^3\\ x_1=1\\ P_1=3p{\left(1-P\right)}^2\end{array}$

More values are mentioned below.

$\begin{array}{c}{x}_3=2\\ P_2=3p^2(1-p)\\ x_4=3\\ P_4=p^3\end{array}$

The expected value of value x is given below.

$\begin{array}{c}E\left(x\right)=\sum _{p^2}{x}_i{p}_i\\ =-6p^2(1-p)+3p{\left(1-p\right)}^2+3p^3\\ =6p^2-6p^3+3p\left(1-2p+p^2\right)+3p^{2}3p\end{array}$

The value of the random variable y is mentioned below.

$\begin{array}{c}{y}_1=0\\ P_1=P^3\\ y_2=1\\ P_2=3p^2(1-p)\end{array}$

More values are mentioned below.

$\begin{array}{c}{y}_3=2\\ P_1=3p{\left(1-p\right)}^2\\ y_4=3\\ P_i={\left(1-p\right)}^3\end{array}$

The expected value of y is given below.

$\begin{array}{c}E\left(y\right)=\sum y_i{p}_i\\ =6p{\left(1-p\right)}^2+3p^2(1-p)+3{\left(1-p\right)}^3\\ =3p^2-3p^3+6p\left(1-2p+p^2\right)+3\left(1-3p+3p-p^3\right)\\ =3-3p\end{array}$

The value is mentioned below.

$\begin{array}{c}{\mu }_{x-y}=E(x-y)\\ =3p-[3-3p]\\ =6p-3\\ =3(2p-1)\end{array}$

Hence, the required value is $3(2p-1)$.