The probability that X can take the values $x_i$ has the following form:

$P\left(X=x_i\right)=\left\{\begin{array}{lll} 0.2 & \text { if } ~~~x_i=0 \\ k x_i & \text { if } ~~~x_i=1 \text { or } 2 \\ k\left(5-x_i\right) & \text { if } ~~~x_i=3 \\ 0 & ~~~~~~\text { otherwise } \end{array}\right.$

The value of P(X = 2) is:

$\frac{8}{25}$

The correct answer is Option (4) → $\frac{8}{25}$

$∑P(x_i)=1$

$⇒0.2+k×1+k×2+k×(5-3)+0=1$

$⇒0.2+k+2k+2k=1$

so $5k=0.8$

so $k=\frac{0.8}{5}=\frac{4}{25}$

so $P(x=2)=2k=\frac{8}{25}$