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Math Help - 3 Probability Questions

  1. #1
    Newbie
    Joined
    Apr 2006
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    3 Probability Questions

    Here are the questions and I will tell you my thought process, you can see where I can go wrong or if I'm approaching it completely wrong.

    Suppose that among the 6000 students at a high school, 1500 are taking honors courses and 1800 prefer watching basketball to watching football. If taking honor courses and preferring basketball are independent, how many students are both taking honors courses and prefer basketball to football?
    A. 300
    B. 330
    C. 450
    D. 825
    E. There is insufficient information to answer this question.

    Here is what I did. P(A)=1500/6000=.25. P(B)=1800/6000=.3.
    P(A u B) = P(A) + P(B) - P(A n B)
    P(A u B) = .3 * . 25 = .075
    Solve and P(A n B) = .475
    .475(6000) = 2850 - obviously wrong

    As reported in the NY Times (Feb 19, 1995, pg. 12), the Russian Health Ministry announced that 1/4 of the country's hospitals had no sewage system and 1/7 had no running water. What is the probability that a Russian hospital will have at least one of these problems...
    a. if the two problems are independent?
    b. if hospitals with a running water problem are a subset of those with a sewage problem?

    A. I did 1/7 + 1/4 = 11/28
    B. P(A|B) = (1/7)(1/4) / (1/4) = 1/7

    11/28, 1/7 is one of the multiple choice answers but I'm not sure if I did that right.

    Suppose that, for any given year, the probabilities that the stock market declines, that women's hemlines are lower, and that both events occur are, respectively .4, .35, .3. Are the two events independent?

    A. Yes, because (.4)(.35) does not equal .3
    B. No, because (.4)(.35) does not equal .3
    C. Yes, because .4 > .35 > .3
    D. No, because .5(.3 + .4) = .35
    E. There is insuffient information to answer this question.

    I know that two events are independent if P(A|B) = P(A).

    P(A|B) = P(A n B) / P(B) = .3 / .35 = .857

    P(A) = .4

    So they are not independent, that limits A, C, E. But neither B or D seems to approach the solution the way I did.
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  2. #2
    Newbie
    Joined
    Mar 2006
    Posts
    8
    Hello, Froggy. Let's begin.
    1) You have confused something. If you need to find the probability that BOTH events A and B happen, you need to find P(A n B). If A and B are independent then P(A n B)=P(A)*P(B). You have already find it. It is equal to .3 * . 25 = .075. So the number of students are both taking honors courses and prefer basketball to football is .075*6000=450

    2) We need to find P(A u B). So it is here where we use the formula P(A u B) = P(A) + P(B) - P(A n B).
    a) If A and B are independent then P(A n B)=P(A)*P(B)=1/4*1/7=1/28. Then
    P(A u B) = P(A) + P(B) - P(A n B)=1/4+1/7-1/28=10/28=5/14
    b) If B is a subset of A then P(A n B)=P(B). Then
    P(A u B) = P(A) + P(B) - P(B)=P(A)=1/4

    3) The events A and B are called independent if P(A n B)=P(A)*P(B). So B is correct.

    Hope this will help you. And one recommendation: learn your textbook more attentively. Good luck.
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