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Homework Help: Throwing Dice and Probability

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A die is rolled three times. What is the probability that you get a larger number each time?

    3. The attempt at a solution

    So I observed that the total number of possibilities for rolling the three dice is, by the Fundamental Rule: [itex](6)(6)(6) = 6^3 = 216[/itex]. I also thought that the number of increasing 3-sequences from the six values on the die is, by applying n-choose-k, [itex](\frac{6}{3}) =20[/itex] (this isn't really a fraction, but I wasn't sure how to format it properly). Thus, the probability of rolling the desired result is [itex]\frac{20}{216}[/itex]. Is that right?
  2. jcsd
  3. Mar 28, 2012 #2
    Correct. Or you could take the long way and count them:


    20 options total
  4. Mar 28, 2012 #3
    Okay, thanks! I also need some guidance with this problem:

    In a draft lottery containing the 366 days of he year (including February 29), what is the probability that the first 180 days drawn are evenly distributed among the 12 months?

    So I thought my first step was to calculate [itex]_{366} C _{180} = \frac{366!}{(180!)(366-180)!}[/itex]. This should give me the total number of combinations of days, right?

    But how do I then incorporate the part of the question that asks about their distribution across the 12 months?
  5. Mar 28, 2012 #4
    I'm not quite sure what that is supposed to mean. Does that mean there are an even amount of the 180 drawings in each month?
  6. Mar 28, 2012 #5
    I think it means to say that, if we select 180 days from the 366 days in the year, what is the probability that we will select 15 days from each of the 12 months (15*12=180)
  7. Mar 28, 2012 #6
    Well think about the months: 1 month has 29 days, 4 have 30 days, and 7 have 31 days

    So you can break it up into parts:

    29C15 [itex]\times[/itex] 30C15 [itex]\times[/itex] ... [itex]\times[/itex] 30C15 [itex]\times[/itex] 31C15 [itex]\times[/itex] ... [itex]\times[/itex] 31C15

    Each of which should be divided by n-15C15:

    366C15, 351C15, 336C15, etc

    Sorry for the multiple edits, I was trying to find the best way to write it.
    Last edited: Mar 28, 2012
  8. Mar 28, 2012 #7
    Hmm. It isn't immediately apparent to me why you divided by [itex]_{n-15} C _{15}[/itex]. Why is that?
  9. Mar 28, 2012 #8
    Because for the total probability of each month, you have 15 less days to choose from.
  10. Mar 28, 2012 #9
    You're dividing the number of ways to choose 15 from each month by the number of ways to choose 15 from the total number of days in the year.
  11. Mar 28, 2012 #10
  12. Mar 28, 2012 #11
    In my first post regarding this problem, I thought it would be a good idea to calculate [itex]_{366} C _{180}[/itex]. I'm stilling thinking that I should be dividing by the total number of possible combinations to get the probability. Does this step come later or not at all?

    Well, I see it doesn't come at all from the link you posted.
  13. Mar 28, 2012 #12

    January: Choose 15 days out of 31 days out of 366
    February: Choose 15 days out of 29 days out of 351

    And so on. Every month's probability removes 15 possible days from the previous amount out of the total 366. And 15 over 12 months amount to 180 days.
  14. Mar 28, 2012 #13
    In the link you posted, the answer seems to indicate that the probability that the first 30 draws contain no days from August or September as [itex]\frac {_{305}C_{30}}{_{305}C_{30}}[/itex]. But isn't this just one? How am I supposed to believe that? I find it hard to believe that the event is guaranteed.

    Maybe if they had written [itex]\frac {_{305}C_{30}}{_{366}C_{30}}[/itex] instead?
  15. Mar 28, 2012 #14
    What are the conditions for part (b)? I didn't read them.
  16. Mar 28, 2012 #15
    The question in my text merely asks "What is the probability that the first 30 days drawn contain none from August or September?" I don't think this can be guaranteed true.
  17. Mar 28, 2012 #16
    You are correct. It would be the number of ways to choose 30 out of (366 - 30 - 31) days, divided by the total number of ways to choose 30 out of 366 days.
  18. Mar 28, 2012 #17
    And BTW that picture I found on another website. I did not make that incorrect solution (gotta protect my reputation lol)
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