Question

The discrete random variable X has the probability function.

Value of X = x	0	1	2	3	4	5	6	7
P(x)	0	k	2k	2k	3k	k2	2k2	7k2 + k

Find k

Answer 1

Since the condition of Probability Mass function

`sum_("i" = 2)^oo` P(x_i) = 1

`sum_("i" = 0)^7` P(x_i) = 1

P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P

P(x = 6) + P(x = 7) = 1

0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1

10k² + 9k – 1 = 0

10k² + 10k – k – 1 = 0

10k(k + 1) – 1(k + 1) = 0

(k + 1)(10k – 1) = 0

k + 1 = 0, 10k – 1 = 0

k = – 1, 10k = 1

k = `1/10`

k = – 1 is not possible