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Stats problem

  1. Jun 3, 2005 #1
    Hi, I need some help with a stats question:

    A and B are two mutually exclusive and collectively exhaustive events. C and D are also mutually exclusive and collectively exhaustive. Determine P(B U D) if P(A) = 0.3, P(C/B) = 0.6, and P(D and A) = 0.20. Can anyone help?
  2. jcsd
  3. Jun 3, 2005 #2
    I don't think you even need P(B | D). P(A) = 0.3 means P(B) = 0.7, and then you have what you need. It might help to draw a diagram.
  4. Jun 3, 2005 #3
    What have u attempted so far? Have u found P(B)?

    -- AI
  5. Jun 3, 2005 #4
    well, I found P(B), which is 0.7. The point where im stuck is finding either P(D) or P(C). Once i get those, I would be able to finish it.

    I was also able to get P(C and B) = 0.42. P(D/A) = 0.67. After that, I cant seem to move any further. Help?
  6. Jun 3, 2005 #5
    I'm not sure why you need P(D and A) = 0.2 or even P(A) = 0.3, unless my definition of "/" is wrong. Is "/" the difference?

    I'm going to use "&" for intersection below. "^c" refers to the complement.

    P(C/B) = 0.6 = P(C & B^c) = P(C & A).

    P(C/B)^c = 0.4 = P((C&A)^c)) = P(C^c U A^c) = P(D U B)

    Am I doing something wrong here? I'm not quite sure why the extra info is needed.
  7. Jun 3, 2005 #6
    I think that / is read as "given," or |. P(C/B) = P(C|B) = P(C & B) / P(B)

    You know that P(A & D) is 0.2. Since A & D and B are mutually exclusive (since A and B are mutually exclusive), what does P(B u (A & D)) equal numerically? And algebraically, what does it simplify to, working on the fact that A and B form a partition?
  8. Jun 3, 2005 #7
    If he indeed meant "|", then he should have used that instead of "/". I think all modern keyboards have that character.
  9. Jun 3, 2005 #8
    sorry that is what i mean. B given A for example. I just found the key. I didnt know the slash had some other meaning. Can u guys explain in words what u mean (referring to first response). I would also like clarification on the second persons response. I didnt really understand either one...
  10. Jun 3, 2005 #9
    i guess jevnal what u wrote wouldnt apply, since I miscommunicated. Bicycletree, ive nver herad of partition or anything like that, can u explain?
  11. Jun 3, 2005 #10
    I'm not sure why you need the conditional probability, actually (if I'm understanding the problem correctly). The other two probabilities seem to be enough information.

    Draw a circle. Divide it in half vertically. The left is A, the right is B. Now, divide the circle in half horizontally. The top is C, the bottom is D. The four quadrants are:

    A & C | B & C
    A & D | B & D

    The total of the four quadrant probabilities sum to 1. Also, the physical size of the quadrants in this diagram have nothing to do with the actual relative magnitudes of the probabilities.

    You want to calculate the total of three of the quadrants (A&D, B&C, B&D). You are given P(A&D). Now you need P(B&D) + P(B&C). That should be pretty easy.
  12. Jun 3, 2005 #11
    I think I understand where you are coming from, but I think the kicker is taht only A + B are mutually exclusive and collectively exhaustive, and then C + D are seperately mutually exclusive and collectively exhaustive. Is that that right approach?
  13. Jun 3, 2005 #12
    How is that inconsistent with the picture I drew?
  14. Jun 3, 2005 #13
    hmm i guess ure right, that does work. But, my question is, based on what I was trying to say earlier, wouldnt the probability of A + B = 1 and C + D = 1, instead of A+B+C+D = 1? I was thinking the diagram should be more like two circles (drawn like a Venn diagram) with a line dividing each in half.
  15. Jun 3, 2005 #14
    It is indeed true that P(A)+P(B) = 1, and P(C) + P(D) = 1. And that is completely consistent with my diagram. My diagram does NOT say that P(A)+P(B)+P(C)+P(D) = 1. It says that P(A&C) + P(A&D) + P(B&C) + P(B&D) = 1.

    All I did was take two separate diagrams and overlay them on each other to show that A can overlap with C and D, and B can overlap with C and D.
  16. Jun 3, 2005 #15
    I would also suggest using a concrete example to make this clearer:

    Start with the population of a city. Pick a person from that population.

    P(A) = probability that person is a male
    P(B) = probability that a person is a female

    P(C) = probability that a person is younger than 30
    P(D) = probability that person is older than or exactly 30.
  17. Jun 3, 2005 #16
    I have A & D, and I can get B & C, but how can I get A & C?
  18. Jun 3, 2005 #17
    i think i might be confused a bit, they want P(B U D). even if I get each quadrant, i wouldnt know the P(D), so how would I approach it?
  19. Jun 3, 2005 #18
    Please read what I wrote more carefully. You don't seem to be. I said you just need P(B&D) + P(B&C). That is very easy to calculate, especially if you examine my diagram and reduce it to a simpler form.
  20. Jun 3, 2005 #19
    I dont understand. Going back to what u said, u say I need the 3 quadrants, fine. But how can I get them? Ive figured out B&C, but still need B&D. Im not seeing how I can easily get that...
  21. Jun 3, 2005 #20
    No - you need the sum of (B&C) and (B&D). You don't need to calculate them separately.

    Going back to my concrete example:

    P(B&C) = probability that person is a female and under the age of 30
    P(B&D) = probability that a person is a female and over or exactly the age of 30

    What is P(B&C) + P(B&D)? Think about it.

    I think you are focusing too much on the formulas, and less on the meaning.
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