Question

The random variable X takes on the values 1, 2 (or) 3 with probabilities

$\frac{2+5P}{5},\frac{1+3P}{5}\text{and}\frac{1.5+2P}{5}$

respectively the values of P and E(X) are respectively

• A. 0.05, 1.87
• B. 1.90, 5.87
• C. 0.05, 1.10
• D. 0.25, 1.40

Answer 1

The correct option is A 0.05, 1.87

Prob. Distribution is

$\begin{array}{cccc}x:& 1& 2& 3\\ p\left(x\right):& \frac{2+5p}{5}& \frac{1+3p}{5}& \frac{1.5+2p}{5}\end{array}$

We know that $\sum p_i=1$

So, $p_1+p_2+p_3=1$

$\frac{2+5P}{5}+\frac{1+3P}{5}+\frac{1.5+2P}{5}=1$

$\Rightarrow P=0.05$

$E\left(x\right)=\sum p_i{x}_i=p_1{x}_1+p_2{x}_2+p_3{x}_3$

$=1\left(\frac{2+5p}{5}\right)+2\left(\frac{1+3p}{5}\right)+3\left(\frac{1.5+2p}{5}\right)$

$=\frac{8.5+0.85}{5}=1.87$