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Defining probability events

  1. Sep 28, 2009 #1
    A box contains one yellow, two red, and three green balls. Two balls are randomly chosen without replacement. Define the following events:

    A: \{ One of the balls is yellow \}
    B: \{ At least one ball is red \}
    C: \{ Both balls are green \}
    D: \{ Both balls are of the same color \}



    3. The attempt at a solution

    Alright, this part is just the setup to the rest of the problem, but I'm still struggling in this area of the class. Can someone try to explain to me how you define the events numerically?
     
  2. jcsd
  3. Sep 28, 2009 #2
    Still haven't been able to figure this out, can anyone help?
     
  4. Sep 28, 2009 #3

    Mark44

    Staff: Mentor

    A, B, C, and D: There are 36 pairs of colors possible. Make a 6 X 6 table. Along the left side, label the rows Y R R G G G, and label the columns the same. In each of the 36 boxes of the table write a pair of letters defined by which row it's in and each column.
    Then for each problem, count the number of boxes that satisfy the particular event. The probability of that event will be the number of boxes that make up the event divided by 36.
     
  5. Sep 28, 2009 #4
    - I would assume that for D, you would ignore the YY column since that is not possible given that only 1 of the balls is yellow? I'm not getting the correct answers for the actual problem portion:

    https://webwork.math.lsu.edu/webwork2_files/tmp/equations/4e/8c463ea6be26c39122f44993cdf9df1.png=[/URL]
    https://webwork.math.lsu.edu/webwork2_files/tmp/equations/b2/d7de36f0f41c31b27bcb1db8fe7cc21.png=[/URL]
    https://webwork.math.lsu.edu/webwork2_files/tmp/equations/a6/92530abeb12d8698db7c64dde491291.png=[/URL]
     
    Last edited by a moderator: Apr 24, 2017
  6. Sep 28, 2009 #5

    Mark44

    Staff: Mentor

    How about put an X in the YY box, since that is obviously not a possibility. That makes 35 possible events, not 36, as I said earlier. How many boxes are there with RR or GG in them?
     
  7. Sep 28, 2009 #6
    RR = 4 and GG = 9
     
  8. Sep 28, 2009 #7

    Mark44

    Staff: Mentor

    So that would make P(two red balls) = 4/35 and P(2 green balls) = 9/35. Do these square with the answers you have?
     
  9. Sep 28, 2009 #8
    I get:
    A = 10/35
    B (compliment) = 15/35
    C = 9/35
    D = 13/35 and D (compliment) = 22/35

    but I can't get the right answer with the problems I posted a few post back.
     
  10. Sep 28, 2009 #9

    Dick

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    Mark44 is suggesting a way to count the draws WITH replacement. Not without replacement. That's different. Let's do A. Either the first ball is yellow or the second ball is yellow. The probability the first ball is yellow is 1/6 since there is one yellow ball and six total. The second ball is automatically not yellow. So that's (1/6). If the second ball is yellow then the first ball must not be yellow. Probability (5/6). The probability of the second ball being yellow is now 1/5 since there is now 1 yellow ball amid 5 nonyellow balls. Total probability (1/6)+(5/6)*(1/5). What's that? Try taking a similar approach to the other ones. Pick one and try it. This is not the only way to do it, but it's one way.
     
    Last edited: Sep 28, 2009
  11. Sep 29, 2009 #10
    - I'm not having much luck with this, I still can't seem to get the right answers. I'm not sure what I'm missing here.
     
  12. Sep 29, 2009 #11

    Dick

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    No way to tell unless you show what you are doing.
     
  13. Sep 29, 2009 #12
    - I'm still trying to completely understand the logic in adding 1/6 to 5/6*1/5. I understand that 1/6 is the probability of the 1st ball being yellow, but I'm not sure if I have the logic with the 2nd ball right. I take it as that there is a 5/6 probability of the 1st ball not being yellow and a 1/5 (one ball is already chosen) probability of the 2nd ball being yellow...so 5/6 and 1/5. So the total probability of one ball being yellow is 1/6+5/6*1/5...is that the right logic?
     
  14. Sep 29, 2009 #13
    I end up getting:

    A = 1/3
    B = 2/6*4/5 + 4/6*2/5 + 2/6*1/5 = 3/5
    C = 3/6*2/5 = 1/5
    D = 2/6*1/5 + 3/6*2/5 = 4/15

    I'm positive C and D are right because the problem involving just those 2 was correct. I don't believe that B is right though, because I can't get the right answers with what I have for the two problems involving B. Can someone check my work? Maybe I'm working the actual problems out wrong?

    https://webwork.math.lsu.edu/webwork2_files/tmp/equations/4e/8c463ea6be26c39122f44993cdf9df1.png
    https://webwork.math.lsu.edu/webwork2_files/tmp/equations/b2/d7de36f0f41c31b27bcb1db8fe7cc21.png
     
    Last edited by a moderator: Apr 24, 2017
  15. Sep 29, 2009 #14

    Dick

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    [/URL]

    There's another way to do B. It's 1-(1st not red)*(2nd not red). 1-(4/6)*(3/5)=3/5. I agree with you on that one.
     
    Last edited by a moderator: Apr 24, 2017
  16. Sep 29, 2009 #15
    Do you not agree with one of the others I have done?
     
  17. Sep 29, 2009 #16

    Dick

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    I didn't say I didn't agree. I haven't checked them all. You know I agree with A. You said you knew C and D were ok. So I just checked B since you were worried about it.
     
  18. Sep 29, 2009 #17
    - Right. So I must be doing my math wrong for the actual problems. For instance,

    https://webwork.math.lsu.edu/webwork2_files/tmp/equations/b2/d7de36f0f41c31b27bcb1db8fe7cc21.png=[/URL] 1 - P(D|B)

    but I'm not getting the correct answer. So I must be doing something wrong in my math. I'm doing P(D|B) like this: P(D|B) = P(DB)/P(D) = P(D)/P(B) , but I think that is probably wrong. Can you help me on this?
     
    Last edited by a moderator: Apr 24, 2017
  19. Sep 29, 2009 #18

    Dick

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    I would if I understood what problem you were trying to solve. P(D given B)=P(D and B)/P(D), if I understand your notation. But that's not equal to P(D)/P(B), is it? Understand, I'm not a master of probability.
     
    Last edited by a moderator: Apr 24, 2017
  20. Sep 29, 2009 #19
    -Ahh, well, I'm going to keep messing with this and see if I can't figure it out. I used the reduction of sample space for another part of the problem in which C was part of the subset of D, so P(D)/P(B) was valid. That won't work with the other 2 problems though. I don't recall ever struggling with anything math related like this before...pretty frustrating.
     
  21. Sep 30, 2009 #20

    Dick

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    Good luck with the notation. You seem to have the probability part down.
     
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