Question

A random variable X has the following probability distribution.

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k2	2k2	7k2 + k

Determine

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

Answer 1

(i) It is known that the sum of probabilities of a probability distribution of random variables is one.

k = − 1 is not possible as the probability of an event is never negative.

`:.k = 1/10`

(ii) P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2)

(iii) P (X > 6) = P (X = 7)

(iv) P (0 < X < 3) = P (X = 1) + P (X = 2)