I have a problem I'm struggling to solve, unfortunately I don't have the answer to check. It states that a number of connection requests $X_t$ arrive at a service provider in an interval t, and $X_t$ is a random variable with a Poisson distribution. The probability of no requests being made in an interval of 10s is 0.001. What"s the probability of 10 or more requests being made in 30s?

A friend of mine tried using 0.999 as the mean value of requests made in 10s (1-0.001), so in 30s lambda would be 3x0.999 = 2.997, but apparently that is wrong.

A tutor tried to explain to me that you have to take the interval in account first, so if 0.001 is the probability of no requests in 10s, 0.003 is the probability of no requests in 30s. So in 30s, the probability of a request being made would be 0.997 (1-0.003), but I can't see how that would be my mean value. Any one care to help?


2 Answers 2


If the number of requests is 10s is $X_{10} \sim \mathsf{Pois}(\lambda_{10}=-\ln(0.001) = 6.908),$ then the number of requests in 30s is $X_{30} \sim \mathsf{Pois}(\lambda_{30}=3(6.908) = 20.72),$ adjusting the rate to the time interval as your tutor says. Also, essentially this much is shown in the answer by @madprob. (+1)

Then you seek $P(X_{30} \ge 10) = 1 - P(X_{30} \le 9)= 0.9967,$ which you can evaluate using computation with the PDF formula of the Poisson distribution (which is a little tedious), by normal approximation (with normal $\mu = 20.72$ and $\sigma = \sqrt{20.72}),$ or using software. Results from R statistical software (in which ppois is a Poisson CDF) are as follows:

1 - ppois(9, 20.72)
## 0.9967306                       # Exact Poisson probability
1 - pnorm(9.5, 20.72, sqrt(20.72))
## 0.9931473                       # Normal aprx with continuity correction

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In this type of question, it is common to assume that $X_{t} \sim \text{Poisson}(\lambda t)$. Note that the average number of connection requests until time $t$ is $\lambda t$, which increases linearly with $t$.

The problem states that $P(X_{10}=0)=0.001$ and asks for $P(X_{30}=0)$. Note that the equation $P(X_{10}=0)=0.001$ determines the value of $\lambda$ above. Indeed, since $X_{10} \sim \text{Poisson}(10\lambda)$, \begin{align*} 0.001 = P(X_{10}=0) = \frac{\exp(-10\lambda)(10\lambda)^{0}}{0!} = \exp(-10\lambda) \end{align*} Therefore, \begin{align*} \lambda &= \frac{-\log(0.001)}{10} \end{align*} Now that you know the value of $\lambda$, you can determine the distribution of $X_{30}$. $X_{30} \sim \text{Poisson}(30\lambda)=\text{Poisson}\left(-3\log(0.001)\right)$. Finally, \begin{align*} P(X_{30}=0) &= \frac{\exp(3\log(0.001))(-3\log(0.001))^{0}}{0!} \\ &= (0.001)^{3} \end{align*} After you do a couple of exercises like this one, you might notice a pattern between $P(X_{t_1}=0)$ and $P(X_{t_2}=0)$. Indeed, it is very suspicious that $P(X_{30}=0)=P(X_{10}=0)^{\frac{30}{10}}$. Following the same steps as above, you might be able to show that, for every $t_1$ and $t_2$, $P(X_{t_2}=0)=P(X_{t_1}=0)^{\frac{t_2}{t_1}}$.

Finally, if you let $\Phi$ to be the cumulative distribution function of a standard normal, \begin{align*} P(X_{30} \geq 10) &= P\left(\frac{X_{30}-E[X_{30}]}{\sqrt{Var[X_{30}]}} \geq \frac{10-E[X_{30}]}{Var[X_{30}]} \right) \\ &= P\left(\frac{X_{30}-E[X_{30}]}{\sqrt{Var[X_{30}]}} \geq \frac{10-30\lambda}{\sqrt{30\lambda}} \right) \\ &\approx 1-\Phi\left(\frac{10-30\lambda}{\sqrt{30\lambda}}\right) \\ &\approx 1-\Phi\left(-2.36\right) \\ &\approx 1-0.009=0.991 \end{align*}

  • $\begingroup$ Thanks for your answer, but the problem asks for P(Xt >= 10), when t = 30s. He gives the probability of no requests being made, but asks for the probability of 10 or more requests being made. Also, your answer would give a probability greater than 1. $\endgroup$
    – Josy Sclei
    Nov 13, 2017 at 21:16
  • $\begingroup$ Ops, I corrected the typo and showed more passages. $\endgroup$
    – madprob
    Nov 13, 2017 at 23:01

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