Question

The discrete random variable X has the probability function

X	1	2	3	4
P(X = x)	k	2k	3k	4k

Show that k = 0 1

Answer 1

P(x = 1) = k

p(x = 2) = 2k

p(x = 3) = 3k

P(x = 4) = 4k

Since P(X = x) is a probability Mass function

`sum_(x = 1)^4` P(X = x) = 1

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

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

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

k + 2k + 3k + 4k = 1

10k = 1

k = `1/10`

∴ k = 0.1