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Poisson Process and probability

  1. Jun 16, 2006 #1
    Suppose the probability is 0.8 that any give person will believe a tale about the transgression of a famous actress.what is the probability that
    (a)the sixth person to hear this tale is the fourth one to believe it?
    (b)the third person to hear this tale is the first one to believe it?

    can anyone help me to solve this question by using poisson process/distribution ???
    Thanks in advance:biggrin:
  2. jcsd
  3. Jun 16, 2006 #2
    The Poisson distribution may not be very helpful in solving this question. Observe that we are not given the mean (a data needed for calculations involving the Poisson distribution), but are told that there is a constant probability of success (where "success" is defined as a random person believing the rumour). Which distribution comes to mind?

    Also, what are your thoughts regarding the question? Part b seems quite easy. Would you like to show some of your working?
  4. Jun 16, 2006 #3
    i m not able to understand how to sort out as the poisson process involves the meu and t and in the question we have probability so it's quite confusing to mee
    how u'll solve the b part ??
    Last edited: Jun 16, 2006
  5. Jun 16, 2006 #4
    I would suggest the Binomial distribution for this question.

    As for part b, for the third person hearing this tale to be the first one to believe it, the first two people who heard the tale must not believe it and the third person must. What is the probability that a person does not believe the tale? What then is the probability that the third person hearing this tale is the first one to believe it?
  6. Jun 16, 2006 #5
    what wud be the value of
    P (x =?)
  7. Jun 16, 2006 #6
    well i did got the answer by my way now u shud tell me is it rite
    the probability of failure is (q=1-p) which is (q=1-0.8=0.2)
    the person who don't belive the prob for them is P(x=who dont belive) (0.2*2=0.04)
    now if we multiply both the probabilities we will get the answer
    P= (0.04*0.8=0.0032)
  8. Jun 17, 2006 #7
    Actually, for part b, the use of the Binomial distribution is optional. But I shall show you how it can be used, and hopefully this will guide you in solving the first part.

    Let X be the random variable for the number of people (out of a total of 2) who believe the tale. Thus, X~B(2, 0.8).
    So, [tex]P(X=0)=(0.8^0)(0.2^2)\left(\begin{array}{cc}2\\0\end{array}\right)[/tex] You will thus get P(X=0)=0.04.

    Since the question specifies that the third person hearing the tale MUST believe it, we do not consider it in the Binomial distribution, as we do not fix the position of success and failure for the Binomial distribution. So, P(third person hearing this tale is the first one to believe it)= [P(X=0)] x 0.8 = 0.032

    Can you solve the first part now?
    Last edited: Jun 17, 2006
  9. Jun 24, 2006 #8
    hii sorry i was out of station and thanks for ur help...i want another favour if u can tell me that is there anysoftware which can help me in generating mathematical symbols or like the one used above i need to if anyone can help me out plzzz?
  10. Jun 24, 2006 #9
    Last edited by a moderator: Apr 22, 2017
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