Question

The p.m.f. of a random variable X is as follows:

P (X = 0) = 5k², P(X = 1) = 1 – 4k, P(X = 2) = 1 – 2k and P(X = x) = 0 for any other value of X. Find k.

Answer 1

Since `sumP` = 1

∴ 5k² + 1 – 4k + 1 – 2k = 1

∴ 5k² – 6k + 1 = 0

∴ 5k² – 5k – k + 1

∴ 5k(k – 1) – 1(k – 1) = 0

∴ (k – 1)(5k – 1) = 0

∴ k = 1 or k = `1/5`

But if k = 1, then P(X = 0) = 5 > 1

Hence k = 1 is discarded

∴ k = `1/5`