The number of jokes a dad tells his children at home per t minutes has a Poisson distribution with a mean of 0.3t. If he gets home at 6.00 pm, what is the probability that his first joke will arrive before 6.30 pm?
I'm not sure how to start this. Any hints?
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You are given that the number of jokes to occur in a time interval $t$ is given by a Poisson distribution. From the derivation of the Poisson distribution, this implies that the probability for a joke event to occur in a small time interval, $\Delta t$ is constant in time, $p = 0.3 \Delta t$, where times are measured in minutes.
The question asks about the time of the next joke. Use the exponential distribution to answer the question. Note that the exponential distribution also applies to situations where the probability of an event to occur is constant in time. It describes the distribution of times until the next event. (This is in contrast to the Poisson distribution which describes the distribution of numbers of events in a fixed time interval.)
You just need to figure out the appropriate "decay constant" (or its inverse, the "rate parameter"). Either integrate the probability density or use the cumulative to work out the answer.
The number of jokes between 6:00 and 6:30 pm is $X \sim \mathsf{Pois}(\lambda = 9).$ If the first joke is before 6.30, then we know that $P(X \ge 1) = 1 -P(X = 0) = 1 - e^{-9}.$
As for the distribution of the waiting time $W$ until the next joke, we can find that the probability of no joke in the interval $(0, t)$ is $e^{-.3t}$ But this is the same as saying $P(W > t) = e^{-.3t}$ or that $$F_W(t) =P(X \le t) = 1 - e^{.3t},$$ for $t > 0.$ So you have the CDF $F_W(t).$ You can differentiate it to get the density function $f_W(t).$ From there you can find the mean time $E(W) = 1/0.3 = 3.333$ until the next joke or the probability (about 0.95) that the next joke will be in 10 minutes or less. The continuous random variable $W$ has an exponential distribution with rate parameter $\lambda = 0.3.$