Question

Let $X_1,X_2\text{and}{X}_3$ be independent and identically distributed random variables with the uniform distribution on [0, 1]. The probability $P(X_1+X_2\le X_3)$ is the largest} is .

• A. 0.16

Answer 1

The correct option is A 0.16

The probability density function of a uniform distribution function on [a, b] is

$f\left(x\right)=(\begin{array}{cc}\frac{1}{(b-a)},& a\le x\le b\\ 0,& \mathrm{otherwise}\end{array}$

$=(\begin{array}{cc}1,& 0\le x\le 1\\ 0,& \mathrm{otherwise}\end{array}$

$E\left(x_1\right)=\frac{a+b}{2}=E\left(X_2\right)=E\left(x_3\right)=\frac{1}{2}$

$V\left(x_1\right)=\frac{(b-a{)}^2}{12}=\frac{1}{12}=V\left(x_2\right)=v\left(x_3\right)$

Let $u=x_1+x_2-x_3$

$\mu =E\left(u\right)=E(x_1+x_2-x_3)$

$=\frac{1}{2}+\frac{1}{2}-\frac{1}{2}=\frac{1}{2}$

$Var\left(u\right)=Var(x_1+x_2-x_2)$

$=Var\left(x_1\right)+Var\left(x_2\right)+Var\left(x_3\right)+0$

$=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{1}{4}$

So $SD,{\sigma }_u=\frac{1}{2}$

So $Z_u=\frac{u-\mu }{{\sigma }_u}=\frac{0-\frac{1}{2}}{\frac{1}{2}}=-1$

Hence

$P(x_1+x_2\le x_3)=P(x_1+x_2-x_3\le 0)$

$=P(u\le 0)$

$=P(z_u\le -1)$

$=P(z_u\ge 1)$

(using symmetry of Gaussian Variable)

$=\frac{1}{2}-P(0<z<1)$

= 0.5 - 0.3413
= 0.1587