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If 30% of students in a physics class also take a calculus course & further assume that P(“earn A in physics”|”taking calculus”) = .4, P(“earn A in physics”|”not taking calculus”) = .1 If a student earns an A in physics what is the probability said student is taking calculus?

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What are you stuck on? –  Daryl Sep 30 '12 at 23:17
    
@Daryl I'm not sure if I'm plugging them in correctly. You don't have to do math but can you set up problem for me? –  kimj Sep 30 '12 at 23:37
    
I think it is ${12\over19}\approx0.631$. Anon did a good job below. –  Daryl Oct 1 '12 at 2:48
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1 Answer

$C$ = "taking calculus"

$C^c$ = "not taking calculus"

$A$ = "earns A in physics"

You want to find $P(C|A)$

Bayes' Theorem = $P(C|A)$ = $\dfrac{P(A|C)P(C)}{P(A)}$

Law of Total Probability: $P(A)$ = $P(A|C)P(C) + P(A|C^c)P(C^c)$

$P(C)$ = .3

$P(C^c)$ = .7

$P(A|C)$ = .4

$P(A|C^c)$ = .1

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I got 0.63157894736. Just wanted to see if you would please work thru so I can compare my answer to yours –  kimj Oct 1 '12 at 1:16
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