Probability homework can you please help me???

Here is the question from my book:

applicants for a temp office at Carter Temp Agency who passed a typing test are placed in suitable positions by Nancy Dwyer and Darla Newberg. Employers who hire temp help through the agency return a card indicating satisfaction or dissatisfaction with the work performance of those hired. From past experience its known that 80% of the employees place by Nancy are rated satisfactory, whereas 70% of those placed by Darla are rated as satisfactory. Darla places 55% of the temporary office help at the agency and Nancy the remaining 45%. If a Carter office worker is rated unsatisfactory, what is the probability that he or she was placed by Darla? Please help me...

You need to use conditional probability, and Bayes' Rule to solve this. I will just use conditional probability. What I will walk you through is a derivation of Bayes' Rule.

For any two events A and B with P(B) 鈮?0 P(A | B) = P(A 鈭?B) / P(B) = P(B | A) * P(A) / P(B)

Let S = satisfactory
Let U = unsatisfactory
Let D = placed by Darla
Let N = placed by Nancy

we know:

P(S | N) = 0.80
P(S | D) = 0.70
P(N) = 0.45
P(D) = 0.55

Find

P( D | U)

= P(U | D) * P(D) / (P(U | D) * P(D) + P(U | N) * P(N))

the numerator is easy enough, the denominator is the law of total probability for finding P(U)

P(U | D) = 1 - P(S | D) = 0.30

= P(U | D) * P(D) / (P(U | D) * P(D) + P(U | N) * P(N))
= (0.30 * 0.55) / (0.30 * 0.55 + 0.20 * 0.45)
= 0.6470588