Job winning probability

[Calculated Risk]

A contractor will bid for two jobs in sequence. She has 0.8 probability of winning the first job. If she wins the first job then she has 0.2 chance of winning the second job; if she loses the first job then she has 0.3 chance of winning the second job. Let X denote the number of jobs that she wins. Find the probability mass function of X.

My solution:
event A: she wins the first job.
event B: she wins the second job.
P(A) = 0.8 => P(A')=0.2
P(B|A) = P(AB)/P(A) = 0.2 => P(AB) =0.16 (Chance of winning both jobs, X=2)
P(B|A') =P(A'B)/P(A') = 0.2
I'm not sure from here I can conclude that P(B) = P(A')x0.2 but this would solve the problem. I know how to continue after this part, I just need to know if it's correct and why it is so.
Thanks.

 

[mikegonz00]

I would suggest using a tree diagram to assign probabilities to each event. You should find that:
P(A)=0.8, P(B|A)=0.2, P(B'|A)=0.8, P(A')=0.2, P(B|A')=0.3, P(B'|A')=0.7

When you're calculating the pmf, you are assigning a probability to each possible value that X can take. X takes values in {0,1,2}. If you're familiar with tree diagrams, just multiply probability (in other words, find the joint probabilities) that correspond to each of these cases. Note that for X = 1 there are two probabilities that must be summed.

 

[Steven G]

She can win:
X=0 jobs in 1 way-- L,L
X=1 job in two ways-- W,L or L,W
X=2 jobs in 1 way-- W,W
Calculate those 4 probabilities and you are done.