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Poisson distribution involving 2 variables

  1. Aug 16, 2014 #1
    1. The problem statement, all variables and given/known data

    A coffee shop sell tea and coffee. The number of cups of coffee sold in a minute can be assumed to be a random poisson variable with mean = 1.5 . The number of cups of tea sold can be assumed to be an independent random variable with mean = 0.5.
    Calculate the probablity that in a period of 3minutes , less than 5 cups of drink were sold.

    2. Relevant equations



    3. The attempt at a solution
    less than 5 cups drink sold= 4 cup + 3 cup + 2 cup +1 cup + 0 cup, ( I let coffee = X , tea = Y) ,
    so i have ( 3X , 1Y ) + ( 3Y , 1X ) +( 2X , 1Y) +( 2Y , 1X) + ( 1X , 1Y) +(2X , 0Y) + (2Y , 0X) + (1X) + (1Y) = 0.877 , please refer to the photo below for the working.

    sorry , since this involve complicated maths formula , so i cant type it one by one and post it here . i attached the working in the photo .
     

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  3. Aug 16, 2014 #2

    Simon Bridge

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    Care with your notation 3X is "3 multiplied into X".
    P(x,y):=P(X=x,Y=y) would be the probability of x coffees and y teas being sold in the same 3min interval.
    By (3X,1Y) I assume you mean 3 coffees and 1 tea, or (x,y)=(3,1).

    I cannot see your reasoning from the sheet of calculations you attached.
    Are tea and coffee sales independent?
     
  4. Aug 17, 2014 #3

    HallsofIvy

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    The problem says "The number of cups of tea sold can be assumed to be an independent random variable". That doesn't say "independent" of what but since "number of cups of coffee sold" is the only other random variable, yes, I believe it is saying that tea and coffee sales are independent. (Which, in real situation, I would not consider likely.)
     
  5. Aug 17, 2014 #4

    Orodruin

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    You *can* do it like that (assuming independent sales). However, splitting it up in all cases is not a priori required. What is the distribution of the sum of two independent Poisson distributions?
     
  6. Aug 17, 2014 #5
    yes , both are independent. can you tell me why i am wrong?
     
  7. Aug 17, 2014 #6

    Simon Bridge

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  8. Aug 17, 2014 #7
    by introducing new variable , i ahve the new mean = 3(1.5+0.50 = 6 , then i plug in the mean into the poisson formua , then i got the ans eventually..
     
  9. Aug 17, 2014 #8
    here's another part of the question .
    In a period of a minuts , exactly 3 cups of drink were sold. Find the probality of that 3 cups sold were coffee ,
    i let X = coffee Y= tea
    I have attached the working in the photo.
    my working is P(3 cups coffee n 0 cup tea ) / P( 3 drinks ) = 0.42

    in the photo , for P( 3 drinks ), i have inserted the value of X and Y and calcultaed outside, the ans is 0.180 the formula for the coffee and tea is shown in the photo. p/s : probability of 3 drinks = P(3X , 0Y) + P(2X , 1Y) +P(1X, 2Y ) + P(3 Y , 0X)= 0.180 P(3 cups coffee n 0 cup tea )= 0.07612


    . then i take P(3 cups coffee n 0 cup tea ) / P( 3 drinks ) = 0.42 , But the ans is 0.442.
     

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  10. Aug 17, 2014 #9

    Ray Vickson

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    Are you saying that for independent Poisson random variables X and Y the total, X+Y, is also a Poisson random variable? That, of course, is a very well-known result and is found in every probability textbook.
     
  11. Aug 17, 2014 #10

    Simon Bridge

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    Please be careful of your notation and comment your work so I can see your reasoning.

    It is not good enough to get the answer - even the correct answer is not good enough - you must also be able to show someone else how you got the answer so that they can easily follow your working.
     
  12. Aug 17, 2014 #11

    Ray Vickson

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    The correct answer is (3/4)^3 = 27/64 = 0.421875, so your answer is OK.

    BTW: The distribution of P(C=k|C+T=n) (C = coffee sales, T = tea sales) is very well known; see almost any textbook and several web pages found from searching on keywords 'poisson distribution'. You will gain a lot more insight if you first evaluate it symbolically and then plug in numbers at the end! This also makes the computations a lot easier.

    Finally: please try to respect PF standards, and post typed versions of your work whenever possible. Your submitted photo is almost useless: it is uncommented, messy and does not show enough intermediate steps. Try to reserve thumbnails for things like drawings and diagrams, etc. Usually I do not look at such thumbnails, and often I cannot even read them at all on some media.
     
  13. Aug 18, 2014 #12
    by introducing new variable = number of cups of coffee and tea sold which is equivalent to sales of coffee + sales of tea. to find the In a period of a minuts , exactly 3 cups of drink were sold. Find the probality of that 3 cups sold were coffee
    i have P(W=3) = ((e^-6)(6^3) )/3! = 0.0892 , then i take 0.07612 divided by 0.0892 , ans = 0.85 , why cant i do in this way?
     
  14. Aug 18, 2014 #13

    Ray Vickson

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    Where does the 0.07612 come from? It looks wrong.
     
  15. Aug 18, 2014 #14
    ( (e^-1.5)(1.5^3)/ 3! ) x ( (e^-0.5)(0.5^0) / 0!) =0.07612
     
  16. Aug 18, 2014 #15

    Ray Vickson

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    Should be
    [tex]P(C = 3,T = 0) = \frac{e^{-c}c^3}{3!} \cdot e^{-t}[/tex]
    where ##c = 3 \times 1.5 = 4.5, t = 3 \times 0.5 = 1.5##.

    That, by the way, is one very good reason for the advice I gave you before (first write things out symbolically, then substitute numbers afterwards). It really does help to avoid blunders, as well as helping to make the work readable.

    Edit: sorry. I see the time period is now 1 minute, not three, so your numerator was correct. Your denominator was wrong, however: it should be
    [tex] P(C+T=3) = \frac{e^{-2} 2^3}{3!}[/tex]
    because C and T are now independent Poisson random variables with means 1.5 and 0.5.

    The answer is still ##(3/4)^3 = 27/64##. This answer would be the same if we asked for the probability of 3 coffees given 3 drinks in 1 year, one microsecond, 1 hour, or any time period! You will see that if you write things out symbolically!
     
    Last edited: Aug 18, 2014
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