Let X be a continuous variable with distribution function F(t) =

{

0, t < 0, 1 − e−2t(1 + 2t + 2t2), t ≥ 0.

The probability P(X ≥ 2|X ≥ 1) equals

(a) 13e−2

(b) 5 13 e−2

(c) 13 5

e−2

(d) 5 13

(e) 1 5

e−2