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Homework Help: 3 Probability Questions

  1. Apr 22, 2006 #1
    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.
  2. jcsd
  3. Apr 23, 2006 #2


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    No, if A and B are independent, the P(AnB)= .3*.25, not P(AuB).
    But what is 0.075*6000?

    It would also be a very good idea to state what you mean by A and B!! I can guess, since I'm trying to help you, that A means "is taking honors courses" and B means "prefers watching basketball to watching football" but if I were a teacher grading this I might not be so nice.

    A) Once again, you've got AnB (A and B both happen) and AuB (either A or B or both happen) reversed. If A and B are independent then P(AnB)= P(A)*P(B). (That's assuming of course, that B means "has a running water problem" and A means "has a sewage problem". You should say that explicitely.)
    B) You didn't answer the question. It didn't ask for P(A|B), it asked for P(AnB). However, since you did that calculation wrong (P(A|B) is not (P(A)P(B)/P(B)), you got the right answer! P(AnB)= P(A|B)*P(B). Since P(B)= 1/7 ("1/7 had no running water") and P(A|B)= 1 ("hospitals with a running water problem are a subset of those with a sewage problem"), P(AnB)= 1/7

  4. Apr 23, 2006 #3
    Okay, so for the first one:

    P(AnB) = .3 * .25 = .075
    .075(6000) = 450

    Still a bit confused on the second one.

    Okay - but the question asked what is the probability that the hospital will have at least one of these problems? What you game me would be solving for both of these problems.

    Third one, is No, b/c (.4)(.35) does not equal .3.
  5. Apr 23, 2006 #4


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    You're right- I misread the question as "both" instead of "at least one".
    Again, letting A= "has a sewage problem" and B= "has a running water problem",
    P(AuB)= P(A)+ P(B)- P(AnB).
    a) A and B are independent. Then P(AnB)= P(A)*P(B)= (1/7)(1/4)= 1/28.
    P(AuB)= 1/7+ 1/4- 1/28.

    b)hospitals with a running water problem are a subset of those with a sewage problem.
    If a hospital has a sewage problem, then it must have a running water problem. That is, P(B|A)= 1 so P(AnB)= P(B|A)P(A)= P(A)= 1/4.
    P(AnB)= 1/4+ 1/7- 1/4.

    which is choice B.
  6. Apr 23, 2006 #5
    These are my options for the question.

    A. 11/28, 1/4
    B. 11/28, 1/7
    C. 9/28, 1/4
    D. 9/28, 1/7
    E. 5/14, 1/4

    Solving for a), the answer in 5/14...
    Solving for b), it's 1/7...
  7. Apr 23, 2006 #6
    Also, there are a few more I'm stuck on now, but I think I got them right...

    Which of the following are true statements?
    I. The probability of an event is always at least 0 and at most 1. (yes)
    II. The probability that an event will happen is always 1 minus the probability that it won't happen. (yes)
    III. If 2 events cannot occur simultaneously, the probability that at least one event will occur is the sum of the respective probabilities of the two events. (yes...I think).
    A. I and II
    B. I and III
    C. II and III
    D. I, II, and III
    E. None of the above gives the complete set of true responses.


    Given that 52% of the U.S. population are female and 15% are older than age 65, can we conclude that (.52)(.15) = 7.8% are women older than age 65?
    A. Yes, by the multiplication rule.
    B. Yes, by conditional probabilities.
    C. Yes, by the law of large numbers.
    D. No, because the events are not independent.
    E. No, because the events are not mutually exclusive.
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