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Homework Help: Probability question on permits

  1. Feb 26, 2010 #1
    A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) in applying for a building permit. Let r.v. X = the number of forms required of the next applicant. The probability that x forms are required is known to be proportional to x; that is, p(x) = cx for x = 1,...,5.

    What is the value of c?

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    I got c = (100 / (1+2+3+4+5)) / 100 = 0.066666667, so that:

    p(1) = 0.07
    p(2) = 0.13
    p(3) = 0.2
    p(4) = 0.27
    p(5) = 0.33

    Is this correct?
     
  2. jcsd
  3. Feb 26, 2010 #2

    tiny-tim

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    Hi BrownianMan! :wink:
    Yes, that's fine. :smile:

    (btw, personally, I'd leave it as c = 1/15, but you'd better check what your professor prefers)
     
  4. Feb 26, 2010 #3
    I have another question:

    A soft-drink machine dispenses only regular Coke and Diet Coke. Sixty percent of all purchases from this machine are diet drinks. The machine currently has ten cans of each type. If 15 customers want to purchase drinks prior to the machine being restocked, what is the probability that each of the 15 is able to purchase the type of drink desired?

    ----------------------------------------------------

    So I wrote down all of the possible combination of preferences - IE 0 want Coke and 15 want Diet Coke, 1 likes Coke and 14 like Diet Coke, etc - and found 16 diferent combinations. Given that there are 20 cans (10 of each) and that 60% of all cans purchased are Diet Coke, I found only one combination in which all 15 people get what they want: 6 like Coke and 9 like Diet Coke. So my answer is 1/16. Is this right?
     
  5. Feb 26, 2010 #4

    vela

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    No. It's not saying that for any set of purchases that exactly 60% of the cans will be Diet Coke, just 60% will be Diet Coke on average. Also the particular combinations of Coke and Diet Coke aren't equally likely, e.g. 8 Coke, 7 Diet Coke sold is less likely than 6 Coke, 9 Diet Coke sold.

    Read up on the binomial distribution and then try again.
     
  6. Feb 27, 2010 #5
    Ok, so what would be the parameters?

    p = 0.6, n = 15, k = 15???
     
  7. Feb 27, 2010 #6

    vela

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    Describe to me what those parameters mean. I ask for two reasons. First, I want make sure the variables are what I think they are as I don't know what conventions your class is using. Second and more important, I want to see if you understand what they are and what they represent in the context of this problem.
     
  8. Feb 27, 2010 #7
    k = # of successes
    n = # of trials
    p = probability of success on each trial
     
  9. Feb 27, 2010 #8

    vela

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    So how do they apply specifically to this problem? What does "a trial" mean in this problem? What is a "success" in this problem?
     
  10. Feb 27, 2010 #9
    I would say that a trial is a person purchasing a drink, and a success is purchasing their preferred drink.
     
  11. Feb 27, 2010 #10

    vela

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    Close. A success has a probability p of occurring, so if p=0.6 is the probability of a person buying Diet Coke, a success would be a person wanting to buy a Diet Coke. A failure would be a person wanting to buy a Coke. So n=15 and k=4, for instance, would correspond to 15 people of which 4 want to buy a Diet Coke and 11 want to buy a Coke.

    Do you see where it goes from here?
     
  12. Feb 27, 2010 #11
    So would the parameters for the probability of all 15 people getting their preferred drink be p=0.6, n=15, k=9? Then the probability would be 0.597??
     
  13. Feb 27, 2010 #12

    vela

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    No. You're making the same mistake you initially made. You're saying the only way for 15 people to get the drink they want from the machine is if exactly 9 want Diet Coke and 6 want Coke. What if 8 wanted Coke and 7 wanted Diet Coke? Wouldn't they all be able to get the drink they wanted?
     
  14. Feb 27, 2010 #13
    Ok, I guess I'm approaching it incorrectly. Am I on the right track, or completely far off? Any hints?
     
  15. Feb 27, 2010 #14

    vela

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    If you have 15 sales, what combinations can be met by the contents of the machine? For example, you can't sell 15 regular Cokes and no Diet Cokes whereas you could sell 6 regular Cokes and 9 Diet Cokes. What are all the other combinations that work?
     
  16. Feb 28, 2010 #15
    Code (Text):
    Coke        Diet Coke
    10               5
     9               6
     8               7
     7               8
     6               9
     5               10
    So six in total.
     
  17. Feb 28, 2010 #16

    vela

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    What's the probability of each combo?
     
  18. Feb 28, 2010 #17
  19. Feb 28, 2010 #18

    vela

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    No. You really have to put a bit more thought into this. You have all the bits and pieces. You need to figure out how to put it all together now.
     
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