Question

A random variable X has the following probability distribution:

Values of X :	0	1	2	3	4	5	6	7	8
P (X) :	a	3a	5a	7a	9a	11a	13a	15a	17a

Determine:
(i) The value of a
(ii) P (X < 3), P (X ≥ 3), P (0 < X < 5).

Answer 1

(i) Since the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

$$\begin{gathered} \Rightarrow a+3a+5a+7a+9a+11a+13a+15a+17a=1 \\ \Rightarrow 81a=1 \\ \Rightarrow a=\frac{1}{81} \end{gathered}$$

(ii) P (X < 3)

$$\begin{gathered} =P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right) \\ =\frac{1}{81}+\frac{3}{81}+\frac{5}{81} \\ =\frac{9}{81} \\ =\frac{1}{9} \end{gathered}$$

P (X ≥ 3)

$$\begin{gathered} P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)+P\left(X=7\right)+P\left(X=8\right) \\ =\frac{7}{81}+\frac{9}{81}+\frac{11}{81}+\frac{13}{81}+\frac{15}{81}+\frac{17}{81} \\ =\frac{72}{81} \\ =\frac{8}{9} \end{gathered}$$

P (0 < X < 5)
$$\begin{gathered} P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right) \\ =\frac{3}{81}+\frac{5}{81}+\frac{7}{81}+\frac{9}{81} \\ =\frac{24}{81} \\ =\frac{8}{27} \end{gathered}$$