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I am quite stuck here, but well I am trying my best.

Question:

The Web page of a teacher receives hits from his students according to a poisson process with rate λ=10 visits/day . Also, his colleagues , are vititng his page with rate 2 visits per/day .

1) What is the probability the visits to exceed 10 in a day?
2) What is the probability no student visits his page into 10 hours interval ?
3) Suppose that every student who visits the site , press the link with teachers publications with chance 2/5. What is the probability into a day that only one student pressed that link?

My attempted Solution:

1) To begin with I turn the days into hours , for my convenience. So I assume that λ1= 10/24 and λ2=2/24=1/12 .

Since the question is not very clear to me... I think that we just need

P((s,s+24))= P((s,s+24))-N(s) >10) =P(N(24)> 10) .

here λ=λ1+λ2 = 1/2 . But, I am not sure how to continue from this point. Should I take some integral from 10 to inf of e^(-λs) * (λs)^x/x! dx ?

Couldn't also I say that its the probability 1-(p(0)+p(1)+...+p(10) )? Although this might take loads of time .

2) So for this question, λ=10/24 . And probably, we need something very very simple ,although for some reason I believe that I am missing something here, but anyway. This: P(0 student visits on (0,10)) so it should be e^100/24 =~ e^-(4.1)

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2 Answers 2

up vote 3 down vote accepted

These look like applications of the formula to compute the probability of exactly $n$ occurrences of a Poisson process in a period during which the expected number of occurrences is $\lambda t$: $\frac{(\lambda t)^n}{n!}e^{-\lambda t}$.

1) there are an expected $12$ visits/day. So the probability of having more than $10$ visits in a day would be the complement of having at most $10$ visits $$ 1-\sum_{k=0}^{10}\frac{12^k}{k!}e^{-12} $$ 2) The expected number of students in $10$ hours would be $10\cdot\frac{10}{24}=\frac{25}{6}$. Thus, the probability of having no students visit the site in a $10$ hour period would be $$ \frac{\left(\frac{25}{6}\right)^{\large0}}{0!}e^{-\frac{25}{6}} $$ 3) The expected number of link presses would be $10\cdot\frac25=4$. Therefore, the probability that exactly $1$ student pressed the link would be $$ \frac{4^1}{1!}e^{-4} $$ Explanation of the Poisson Expectation in 3)

Suppose each event of a Poisson process with an expectation of $\lambda t$ events in time $t$ has a probability $p$ of being special. Then the special events constitute a Poisson process with expectation $\lambda pt$ events in time $t$. In the case in question $\lambda t=10$ and $p=\frac25$ so that $\lambda pt=4$.

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I agree , with the first and 2nd solution , because its exactly like my approach , but could you possibly explain the 3rd part bit more ? I seems that your approach is different to Hans Engler . Thank you. –  Thomas Hrksn Dec 11 '12 at 23:35
    
@ThomasHrksn: I have added some explanation. Let me know if that doesn't explain things. –  robjohn Dec 12 '12 at 0:09
    
Thanks its clear to me now. –  Thomas Hrksn Dec 12 '12 at 13:13

Let $S_t$ be the process that counts the number of student visits up to time $t$, let $C_t$ be the processor counts the number of colleague visits up to time $t$, let $A_t$ be the number of all visits up to time $t$.

You do not need to change the timescale, so just keep it in days.

Then you have assumptions about $S_t$ and about $C_t$. Therefore, you can also say something about $A_t$ assuming that the two processes are independent.

Problem 1 is about the process $A_t$. Problem 2 is about the process $S_t$. Problem 3 is about something called a Poisson-binomial random variable: You observe $S_1$ visits = $S_1$ trials. That's a random integer whose distribution you know. Each trial results in a success (click through to the page with teacher publications) with probability $p = 2/5$. Let $X$ be the number of successes. You want to know the probability that $X = 1$. It is going to be an infinite series which you should however be able to sum.

To get you started, think about it this way. Suppose $p_k$ is the probability that in $k$ visits, there is exactly one click through. Write down the formula for $p_k$, using your knowledge of binomial distributions. Then by the law of total probability,

$$P(X = 1) = P(S_1 = 1)\cdot p_1 + P(S_1 = 2)\cdot p_2 + \dots \, . $$

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