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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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    I think it is ${12\over19}\approx0.631$. Anon did a good job below.2012-10-01

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$C$ = "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 yours2012-10-01