Quant - Total Probability question HELP please!

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Greetings all,
I’m having difficulty understanding a problem in the CFAI material on total probability. I understand the formula, or at least I thought I did, but I’m getting caught on a certain portion. Thank you in advance for your thorough assistance!
so Total Prob rule is p(a)=p(a/s)*p(s)+the complement of those terms - I’ve got that!
in this specific problem, p(a)=.55 p(s)=.55 p(a/s^c)=.40 p(s^c)=.45
in this case, the unknown variable is p(a/s). Why is solving that expression: p(a/s)=[.55-.40(.45)]/.55= .673
more specifically, why would you subtract .55 from the p(a/s^c)*p(s^c) term?
 
Simple algebra:
P(A) = P(A|S)P(S) + P(A|S´)P(S´)
P(A) − P(A|S´)P(S´) = P(A|S)P(S)
P(A|S) = [P(A) − P(A|S´)P(S´)] / P(S)
 
Although I certainly appreciate your help, s2000magician, regurgitating the solution doesn’t explain the ‘simple algebra’ you mentioned. Can anyone else elaborate on the ‘simple algebra’ and why p(a) is subtracted from the complement?
 
p(A)=.55, p(S)=0.55, p(A|S’)=0.40, p(S’)=.45 (Given), p(A|S)= ?
From Total Probability Rule:
p(A) = p(A|S) x p(S) + p(A|S’)x p(S’)
Now substitute the inputs given to you:
0.55 = p(A|S) x 0.55 + 0.40 x 0.45
0.55 = 0.55 x p(A|S) + 0.18
0.55 - 0.18 = 0.55 p(A|S)
0.37 = 0.55 p(A|S)
p(A|S) = 0.37/0.55 = 0.6727 = 0.673 (3 decimal places)
 
You’re not subtracting P(A) from anything.
P(A) has two parts: P(A&S) and P(A&S´):
P(A) = P(A&S) + P(A&S´)
Subtracting P(A&S´) from both sides (that’s what you’re subtracting, you’re not subtracting P(A)) gives:
P(A) − P(A&S´) = P(A&S)
Now we use two facts: P(A&S) = P(A|S)P(S), and P(A&S´) = P(A|S´)P(S´), to get:
P(A) − P(A|S´)P(S´) = P(A|S)P(S)
Finally, because we want to solve for P(A|S), we divide both sides by P(S) to get:
[P(A) − P(A|S´)P(S´)] / P(S) = P(A|S)
Maybe putting an example into words will help. Imagine you know that the probability of a car being red is 10%. There are red Ferraris and red non-Ferraris. Suppose that the probability of a car being red but not a Ferrari is 9%. We subtract 9% from 10%: the probability of a car being red and a Ferrari is 1%. Suppose that 1.25% of all cars are Ferraris. Then the probability of a car being red given that it’s a Ferrari is 1% / 1.25% = 80%.
So the reason we subtract P(A&S´) from P(A) – we subtract 9% from 10% – is to get to P(A&S): the probability that a car is both red and a Ferrari.
Note that we didn’t subtract P(A) – 10% – from anything; we subtracted 9% from 10%.
 
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