Question

The probability distribution function of a random variable X is given by

xi :	0	1	2
pi :	3c3	4c − 10c2	5c − 1

where c > 0
Find: (i) c (ii) P (X < 2) (iii) P (1 < X ≤ 2)

Answer 1

(i) We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) = 1

$$\begin{gathered} \Rightarrow 3c^3+4c-10c^2+5c-1=1 \\ \Rightarrow 3c^3-10c^2+9c-2=0 \\ \Rightarrow \left(c-1\right)\left(3c^2-7c+2\right)=0 \\ \Rightarrow \left(c-1\right)\left(3c-1\right)\left(c-2\right)=0 \\ \Rightarrow c=\frac{1}{3},1,2 \\ \left(\mathrm{Neglecting}1\mathrm{and}2\mathrm{as}\mathrm{individual}\mathrm{probability}\mathrm{should}\mathrm{not}\mathrm{be}\mathrm{greater}\mathrm{than}\mathrm{one}\right) \end{gathered}$$

(ii) P (X < 2)

$$\begin{gathered} =P\left(X=0\right)+P\left(X=1\right) \\ =3c^3+4c-10c^2 \\ =\frac{1}{9}+\frac{4}{3}-\frac{10}{9} \\ =\frac{1+12-10}{9} \\ =\frac{3}{9} \\ =\frac{1}{3} \end{gathered}$$

(iii) P (1 < X ≤ 2)

$$\begin{gathered} =P\left(X=2\right) \\ =5c-1 \\ =\frac{5}{3}-1 \\ =\frac{5-3}{3} \\ =\frac{2}{3} \end{gathered}$$