91Ó°ÊÓ

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.

$\begin{array}{c}{\mathrm{μ}}_{x−v}=E(x−y)\\ =B\left(x\right)−B\left(y\right)\end{array}$

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.

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

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.

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

The value is mentioned below.

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

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