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Problem in probability

  1. Sep 13, 2015 #1
    1. The problem statement, all variables and given/known data
    An imaginary particle can decay in 0, 1 or 2 particles of the same kind with probabilities 1/4, 1/2 and 1/4, respectively. Beginning with one particle, we denote xi the number of particles in the ith generation. Determine:

    a. P(x2 >0)
    b. The probability that x1 = 2, given that x2 = 1.


    2. Relevant equations
    let say event A : the particle decay to 0 particle P(A)=1/4
    event B: the particle decay to 1 particle P(B)=1/2
    event C: the particle decay to 2 particles P(C)=1/4

    the conditional probability :
    P(A|B)= P(A∩B) /P(B)


    3. The attempt at a solution
    For part a:

    I am not sure that I understand the question very well .. is it mean by P(x2 >0) the second generation (x2) can decay to 1 or 2 particles .. if yes shall I do this:
    P(B∩C) =P(B)×P(C)
    =1/2 × 1/4 = 1/8 =13%

    For part b:

    P(x1|x2) = ?
    is it equal to P(B|C) ??


    please give me a hint .. I need to solve this by hand and then I can compute it

    thank you so much :)
     
  2. jcsd
  3. Sep 13, 2015 #2

    PeroK

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    I would say this is an ideal problem to use a probability tree. Your notation using A, B and C, doesn't really help you given that there are different generations of these events.

    I think you've misinterpreted the problem. The initial particle must be the 0th generation (I assume). So, the first generation is what it decays into.
     
  4. Sep 13, 2015 #3
    https://fly.cloud.photobox.com/2984919587ea968d6b9e99fdd1b96c87d958717a3c978173f83c89a622c049a69cc2da81.jpg [Broken]

    is this correct ?
    I think I should not calculate the zero particle ,, right ?
     
    Last edited by a moderator: May 7, 2017
  5. Sep 13, 2015 #4

    PeroK

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    That's a good approach. Remember that if 2 particles are created in the first generation, then both of these will decay independently in the second generation. So, you need a bit more work on the case where ##x1 = 2##
     
  6. Sep 13, 2015 #5
    how about this ?
    and for part b :
    P= 25%

    PeroK .. thank you very much :)
     

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  7. Sep 13, 2015 #6

    PeroK

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    I'm not sure how you get ##P(x2 > 0) = 1##

    There are three ways you can end up with 0 particles:

    ##x1 = 0##
    ##x1 = 1## and ##x2 = 0##
    ##x1 = 2## and ##x2 = 0## (for both particles)

    For part b, how do you get 25%?
     
  8. Sep 13, 2015 #7
    well, the problem ask for the probability of getting number of particles in the second generation more than zero ... not equal to zero right ?
    so I sum up all the probabilities of getting 1 particle and 2 particles in second generation .. I will highlight the path in the attached photo
    I get P=1 when I sum all the probabilities even with getting zero particles .. but now if I remove them I get P=0.75
     

    Attached Files:

  9. Sep 14, 2015 #8

    PeroK

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    ##P(x1 = 0)## = 0.25

    Hence ##P(x2 = 0) > 0.25##

    You must have made a mistake adding up.
     
  10. Sep 14, 2015 #9
    why you are looking for the probability of getting zero particle ?

    for part b .. I would say that when x1=2 and x2=1
    p= (0.25 * 0.5) +(0.25 *0.5) = 0.25

    right ?
     
  11. Sep 14, 2015 #10

    PeroK

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    I don't understand that calculation. The first thing you need is ##P(x2 = 1)##. What do you get for that?
     
  12. Sep 14, 2015 #11
    is the tree that I made wrong ?

    b. The probability that x1 = 2, given that x2 = 1.

    I will explain the calculations:
    p(x1=2) = 0.25
    p(x2=1)= 0.5

    but because the first generation get 2 particles I will multiply it by 2
    2(0.25*0.5) = 0.25

    where is the mistake exactly ?
     
  13. Sep 14, 2015 #12

    PeroK

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    That's just not the correct approach. And, when you did your tree you forgot to combine probabilities in the case where x1 = 2.

    Given x1 = 2, you have:

    ##P(x2 = 0) = (1/4)(1/4) = 1/16##

    ##P(x2 = 1) = 2(1/4)(1/2) = 1/4##

    Etc. All the way to ##P(x2 = 4)##

    In the tree, you need to factor in the probability of 1/4 that ##P(x1 = 2)##

    In any case, ##P(x2 = 1) \ne 0.5##
     
  14. Sep 14, 2015 #13
    cool !
    that take us to say p(x2=1) = 1/4

    is not like this ?
     
  15. Sep 14, 2015 #14

    PeroK

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    That's not right. There is a formula for conditional probabilities, but let me show you how I think about it. In this case we want ##P(x1 = 2 | x2 = 1)##

    First, we identify all the tree branches that result in ##x2 = 1##. There are two of them:

    1) x1 = 1 -> x2 = 1 (this has a probability of 1/4 = 4/16).

    2) x1 = 2 -> x2 = 1 (this has a probability of 1/16)

    Every 16 times we do the experiment, we wil get x2 = 1 on average 5 times: 4 times from the first branch (where x1 =1) and 1 time from the second branch (where x1 = 2).

    So, given that x2 = 1, the probability is 4/5 that x1 =1 and only 1/5 that x1 = 2.

    Note that in this case, the conditional probability was going "backwards". The question is not: given that x1 = 2, what is the probability that x2 = 1. That is simpler, with the conditional probability going "forwards".

    You have to be able to think though conditional probabilities both ways, as it were. For example, here are two different questions:

    a) Given that it's raining today, what's the probability that it will be raining tomorrow.

    b) Given that it's raining today, what is the probability that it was raining yesterday (assumning you don't know!).
     
  16. Sep 14, 2015 #15
    sorry, I forget the conditional probability :(

    P(x1=2|x2=1) =P(x1=2 ∩x2=1) / P(x2=1)

    P(x1=2 ∩x2=1) = 1/16
    P(x2=1)= 5/15

    P(x1=2|x2=1) = 1/5

    correct?

    ________________
    you divide by two instead of multiply !!
    I was double it and get P= 1/4 instead of 1/16 !!
     
  17. Sep 14, 2015 #16

    PeroK

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    I calculated the total probability of that sequence of events. You were calculating the conditional probability of the 2nd event given the first. It's important to understand which one is needed.

    You need a new problem now to test whether you understand these sort of problems. You've got the right answer now, but after a lot of attempts!
     
  18. Sep 14, 2015 #17
    thank you very much
    actually I was working on a computing and I forget some details on the statistical basics .. I didn't take a course on the probabilities

    tank you :)
     
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