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Math Help - hmwk problem

  1. #1
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    hmwk problem

    Students who read their statistics text have a .98 probability of passing the course. Those who don't have a .85 probability of failing. If 70% of the students do the readings, then:

    a) What is the probability that a student will pass the course?

    b) If a student were to pass, what would the probability have been that he or she did the readings?

    a) .371
    b) .398

    __________________________________________________ ________
    This is what I have come up with so far:

    Let P = pass course and T= read text

    (P|T) = 0.98 so P(P'|T) = 0.02
    (P'|T') = 0.85 so (P|T') = 0.15
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by swoopesjr01
    Students who read their statistics text have a .98 probability of passing the course. Those who don't have a .85 probability of failing. If 70% of the students do the readings, then:

    a) What is the probability that a student will pass the course?

    b) If a student were to pass, what would the probability have been that he or she did the readings?

    a) .371
    b) .398

    __________________________________________________ ________
    This is what I have come up with so far:

    Let P = pass course and T= read text

    (P|T) = 0.98 so P(P'|T) = 0.02
    (P'|T') = 0.85 so (P|T') = 0.15
    a)

    <br />
p(P)=p(P|T)p(T)+p(P|T')p(T')<br />

    so:

    <br />
p(P)=0.98\times 0.70+0.15 \times 0.30=0.73<br />

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by swoopesjr01
    Students who read their statistics text have a .98 probability of passing the course. Those who don't have a .85 probability of failing. If 70% of the students do the readings, then:

    a) What is the probability that a student will pass the course?

    b) If a student were to pass, what would the probability have been that he or she did the readings?

    a) .371
    b) .398

    __________________________________________________ ________
    This is what I have come up with so far:

    Let P = pass course and T= read text

    (P|T) = 0.98 so P(P'|T) = 0.02
    (P'|T') = 0.85 so (P|T') = 0.15
    b) Here we need what I think of as 1/2Bayes' Theorem:

    <br />
p(T|P)p(P)=p(T \mbox{ and }P)<br />

    Now:

    <br />
p(T \mbox{ and }P)=0.7 \times 0.98=0.686<br />

    so:

    <br />
p(T|P)=p(T \mbox{ and }P)/p(P)=0.686/0.73\approx 0.940<br />

    RonL
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