cumulative distribution function

Question

Consider a random variable X whose distribution function (cdf) is given by:

FX(x) =

0 if x < −2

0.1 if − 2 ≤ x < 1.1

0.3 if 1.1 ≤ x < 2

0.6 if 2 ≤ x < 3

1 if x ≥ 3

(a) Give the probability mass function, p(x), of X, explicitly.

(b) Compute P(2 < X < 3).

(c) Compute P(X ≥ 3).

(d) Compute P(X ≥ 3 | X ≥ 0).

 

I know that the answer for a is:

Pr(x) = .1 if x = -2

           .2 if x = 1.1

           .3 if x= 2

           .4 if x = 3

           0 if otherwise

b is 0

c is .4

d is .444 but i dont understand how this is the answer... I understand everything else

Answer 1

First of all, good job on parts (a), (b) and (c)!  

 

Part (d) is asking for the conditional probability that X≥3, given that X≥0.  The formula for conditional probability is  

 

P(A|B) = P(A∩B) / P(B)  

 

The probability that X≥3 and X≥0 is the same as P(X≥3), which is simply 0.4 by part (c).  

The probability that X≥0 is 0.9 since the only value of X with non-zero probability that is less than 0 is X = -2 with probability 0.1.  

 

So, P(X≥3|X≥0) = 0.4/0.9 = 4/9 ≅ 0.444.