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)
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