A biased dice is thrown once. If X denotes the number appearing on it…

From distribution

$∑P(x) = 1$

$k + \frac{k}{2}+2k+8k^2+1-5k+\frac{k}{2}=1$

so $8k^2-k=0$

$8k^2=k$

so $k=0$ or $k=\frac{1}{8}$

A. $P(X = 3) = 2k$

B. $P(X ≤ 2) = k + \frac{k}{2} = \frac{3k}{2}$

C. $P(X ≥ 5) = 1 - 5k + \frac{k}{2} = 1 - \frac{9k}{2}$

D. $P(X = 4) = 8k^2$

E. $P(X = 1) + P(X = 5) = k + 1 - 5k = 1 - 4k$

for $k=0$

$A=0,B=0,C=0,D=0,E=0$ (Neglected)

for $k=\frac{1}{8}$

$A=\frac{1}{4}, B=\frac{3}{16},C=\frac{7}{16},D=\frac{1}{8},E=\frac{1}{2}$

$A=\frac{4}{16}, B=\frac{3}{16},C=\frac{7}{16},D=\frac{2}{16},E=\frac{8}{16}$

so $E>C>A>B>D$