A discrete random variable X takes the values 0, 1, 2, 3, 4 and its mean is 1.6. If P(X = 1) = 0.4, P(X = 4) = P(X = 2) and P(X = 3) = 2 P(X = 2), then P(X = 0) is:

0.2

The correct answer is Option (1) - 0.2

$\text{Let } P(X=2)=p$

$P(X=4)=p,\;\; P(X=3)=2p,\;\; P(X=1)=0.4$

$P(X=0)=1-(0.4+p+2p+p)=0.6-4p$

$E(X)=1.6$

$0\cdot P(X=0)+1(0.4)+2p+3(2p)+4p=1.6$

$0.4+2p+6p+4p=1.6$

$0.4+12p=1.6$

$12p=1.2 \Rightarrow p=0.1$

$P(X=0)=0.6-4(0.1)=0.2$

$P(X=0)=0.2$