Question

The range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c³, P(X = 1) = 4c – 10c², P(X = 2) = 5c – 1

(i) Find the value of c 

(ii)P(X < 1), P(1 < X ≤ 2) and P(0 < X ≤3)

Answer 1

P(X = 0) + P(X = 1) + P(X = 2) = 1

3c³ + 4c – 10c² + 5c – 1 = 1

3c³ – 10c² + 9c – 2 = 0 

c = 1

satisfy this equation

c = 1 ⇒ P(X = 0) = 3

which is not possible

Dividing with c – 1, we get 3c² – 7c + 2 = 0

(c – 2) (3c – 1) = 0

c = 2 or c = 1/3

c = 2 ⇒ P(X = 0) = 3.2³ = 24

which is not possible