Suppose that the random variable X is defined as X=0.6X1+0.3X2+0.1X3, where the probability distributions of...

Step 1
Given:
${\displaystyle X=0.6X_1+0.3X_2+0.1X_3}$
Where all three are independent
$$f_{X1}(x)=\left(\begin{array}{c}5\\ x\end{array}\right){0.1}^x\times {0.9}^{5-x}$$
$$f_{X2}(x)=\left(\begin{array}{c}5\\ x\end{array}\right){0.4}^x\times {0.6}^{5-x}$$
$$f_{X3}(3)=\left(\begin{array}{c}5\\ x\end{array}\right){0.6}^x\times {0.4}^{5-x}$$
${\displaystyle x=0,1,2,3,4,5}$
Step 2
It can be seen that all the three distributions follows binomial distribution
So, Mean of binomial is
${\displaystyle n\times p}$
Variance of binomial is
${\displaystyle n\times p\times q}$
So, ${\displaystyle X_1\sim B\in (5,0.1)}$
${\displaystyle E\left[X_1\right]=5\times 0.1=0.5}$
${\displaystyle Var\left[X_1\right]=5\times 0.1\times 0.9=0.45}$
${\displaystyle X_2\sim B\in (5,0.4)}$
${\displaystyle E\left[X_2\right]=5\times 0.4=2}$
${\displaystyle Var\left[X_2\right]=5\times 0.4\times 0.6=1.2}$
${\displaystyle X_3\sim B\in (5,0.6)}$
${\displaystyle E\left[X_3\right]=5\times 0.6=3}$
${\displaystyle Var\left[X_3\right]=5\times 0.6\times 0.4=1.2}$
Step 3
a) Mean of X
${\displaystyle E\left[X\right]=0.6\times E\left[X_1\right]+0.3\times E\left[X_2\right]+0.1\times E\left[X_3\right]}$
${\displaystyle E\left[X\right]=0.6\times 0.5+0.3\times 2+0.1\times 3}$
${\displaystyle E\left[X\right]=1.2}$
Hence, the mean is 1.2
Step 4
b) If the events are independent, then the variance is
${\displaystyle Var[A+B]=Var\left[A\right]+Var\left[B\right]}$
Here, all are independent, so the variance of X can be calculated as follows
${\displaystyle Var\left[X\right]={0.6}^2\times Var\left[X_1\right]+{0.3}^2\times Var\left[X_2\right]+{0.1}^2\times Var\left[X_3\right]}$
${\displaystyle Var\left[X\right]={0.6}^2\times 0.45+{0.3}^2\times 1.2+{0.1}^2\times 1.2}$