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Conditional Probability (with integrals)

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose a person's score X on a math aptitude test is a number between 0 and 1, and their score Y on a music aptitude test is also between 0 and 1. Suppose further that in the population of all college students in Canada, the scores X and Y are distributed according to the joint probability density function

    [tex]f(x,y) = \frac{2}{5}(2x + 3y) if 0 \leq x \leq 1 and 0 \leq y \leq 1[/tex]

    0 otherwise.

    a) What proportion of college students obtain a score greater than 0.8 on their math test?
    b) If a randomly selected student's score on the music test is 0.3, what is the probability that this student's score on the math test will be greater than 0.8?

    2. Relevant equations

    3. The attempt at a solution


    [tex]P (X > 0.8) = \int^1_0 \int^1_{0.8} \frac{2}{5}(2x + 3y) dx dy [/tex]


    [tex]P(A|B) = \frac{P(A \cap B)}{P(B)} \rightarrow P (X > 0.8 | Y = 0.3) = \frac{P((X>0.8)\cap(Y=0.3))}{P(Y = 0.3)}= \frac{\int^1_{0.8} \frac{2}{5}(2x + 3(1)) dx }{0.3}[/tex]

    Did I set up both questions correctly?
  2. jcsd
  3. Mar 19, 2009 #2


    Staff: Mentor

    The a part looks to be set up correctly, but something's bothering me about the b part. Isn't the probability of a single value zero? IOW, doesn't P(X = 0.3) = 0? Are you sure you have the wording of the problem exactly right? If the wording is "at least 0.3" or "below 0.3" that would make a difference.
  4. Mar 19, 2009 #3


    Staff: Mentor

    I've thought about this some more. I was assuming that this distribution was continuous, in which case P(X = 0.3) is in fact zero. If, OTOH, the distribution is discrete and uniformly distributed, then P(X = 0.3) is nonzero. If so, and assuming that the only scores possible are {0, .1, .2, ..., .9, 1}, then the probability of any one of these values is 1/11 [itex]\approx [/itex] .0909.
  5. Mar 19, 2009 #4
    I tripled checked, the wording is correct.

    So what you are saying is that P(Y=0.3) is 1/11 due to the fact that there are 11 choices (0, .1, .2, .3, .4, .5, .6, .7, .8, .9, 1) and all of them have an equal chance of occuring.

    [tex]P(A|B) = \frac{P(A \cap B)}{P(B)} \rightarrow P (X > 0.8 | Y = 0.3) = \frac{P((X>0.8)\cap(Y=0.3))}{P(Y = 0.3)}= \frac{\int^1_{0.8} \frac{2}{5}(2x + 3(0.3)) dx}{1/11}[/tex]
    Last edited: Mar 19, 2009
  6. Mar 19, 2009 #5


    Staff: Mentor

    Yep, that's exactly what I'm saying. In the absence of any clarifying information, I don't know what else to think.

    The assumption that the scores are discrete rather than continuous impacts how probabilities are calculated. Instead of using an integral, as is the case for continuous random variables, the thing to do, would be to use a summation, which is what is done for discrete random variables. So P(X > 0.8) = P(X = 0.9) + P(X = 1.0). For each of these you would use your pdf with the appropriate x and y values.

    When you hand in this problem (if that's what you do with them), it would be a good idea to state your assumptions--that the only possible scores are 0, 0.1, 0.2, and so on, and that they are uniformly distributed.

    That's what I think, and that's the way I would tackle this problem. I could be wrong, and welcome any correction to my thinking.
  7. Mar 20, 2009 #6
    I asked my professor, and he said the scores are continous, and not discrete. This is what I thought of:

    [tex]P(X > 0.8 \cap Y = 0.3) = \int^1_0 \frac{2}{5}(2x + 3(Y = 0.3)) dx = \int^1_0 \frac{2}{5}(2x + 3(0.3)) dx[/tex]

    this way, why is no longer a variable, but a constant with a value of 0.3.
  8. Mar 20, 2009 #7


    Staff: Mentor

    That seems reasonable on the surface, but if you do the integration, you get .76. This doesn't seem like a plausible result; namely that more than 3/4 of the students got a score of 0.3 on the music test, and a score over 0.8 on the math test.
  9. Mar 20, 2009 #8
    Maybe all Canadian college students are good at math but suck at music? (Seriously doubt that).

    For a continous case, that's all I could come up with. For all I know, maybe my prof chose those those numbers at random.
  10. Mar 20, 2009 #9


    Staff: Mentor

    Things also don't make sense to me from a geometric standpoint. With a single continuous random variable, the P(X = a) is zero, and P(a < X < b) is the area under the pdf between a and b. For your pdf, the probability of a point is zero, and the probability along a line segment is also zero. An even with nonzero probability would represent some two-dimensional region in the X-Y plane, and the probability of this event could be thought of as the volume under the pdf curve and above the event region.

    So I don't have any more suggestions. When you find out the solution, please post it.
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