Conditional Probability question

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Homework Statement


Students who actively participate in class are 4 times more likely to get a good grade than those who don't. 15% of students actively participate in class; 20% of students get a good grade. A person did not participate in class, what is the probability that he got a good grade?


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The Attempt at a Solution



Well, I define P(G): student got a good grade; P(P): student participates. Obviously, P(G) = .2 and P(P) = .15. I also know that P(G|P) = 4P(G| not P). I am looking for the conditional probability P(G| not P). I would imagine that I need to find P(G|P) / 4, but this would mean that I need [itex] P(G \cap P) P(P) [/itex] and alas, I do not believe that I have [itex] P(G \cap P) [/itex]. ANy ideas? Anything I am missing? Thanks!
 

Answers and Replies

  • #2
jbunniii
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Note that

[tex]P(G) = P(G \cap P) + P(G \cap P^c) = P(G|P)P(P) + P(G|P^c) P(P^c)[/tex]

You know everything except [itex]P(G|P)[/itex] and [itex]P(G|P^c)[/itex], and you know a relation between those two which should allow you to solve for [itex]P(G|P)[/itex].
 
  • #3
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I must be doing something wrong... I have

[itex].2 = (.15)P(G|P) + (.85)P(G|P^c) = (.15)P(G|P^c)/4 + (.85)P(G|P^c)[/itex]

and upon simplification I get [itex]P(G|P^c) = .225[/itex], but the answer is supposed to be .16.
 
  • #4
jbunniii
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Here is what I get. Write [itex]x = P(G | P^c)[/itex]. Then

[tex]0.2 = (0.15) (4x) + (0.85) x = 1.45x[/tex]

so

[tex]x = 0.2 / 1.45 = 0.1379...[/tex]

which doesn't match either of your answers.

P.S. You know .225 has to be wrong, since P(good grade) = 0.2 and surely the probability shouldn't go up for students who don't participate!
 
  • #5
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Thanks for all the help! It wouldn't be the first typo in the notes...
 

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