Probability with Poisson Distribution - find the probability of 10-min event based on 30-mins probability

Question

I need your assistance with the following question:

The probability of a bus arriving at a given station over a 30-minute span of time at least once is 0.9. If you arrive with only 10 minutes to spare, what are the chances a bus arrives in this time span? (assuming the probability is constant throughout the time)

I assume I should use the Poisson distribution, so I did the following:

$$P(x\ge1)=1-P(x<1)=1-P(x=0)=0.9 $$ $$1-\frac{e^{-\lambda}\lambda^0}{0!}=0.9$$ $$\lambda=2.302585$$

But now what should I do with this $\lambda$? How can I proceed?

Thanks!

Answer 1

Hint: $\lambda$ is the expected number of buses in the 30 minute timeframe. So how do you calculate the probability that no busses appear in a 10 minute timeframe?

Answer 2

There is one approach to this problem that does not make use of the Poisson distribution. Consider three intervals of ten minutes each. Now, since the probability of arrival is uniform, we can say that the bus arriving within any one of these ten minute intervals is equal , let this probability be p. Since probability of the bus not showing up at all is 0.1, we have, (1-p)^3 = 0.1, which gives 1-p = 0.464, which gives p = 0.536.

To understand why some equations (like p(1-p)^2 = 0.3) don't give the correct value of p, read the comment by Mick A in this discussion : Probability of crossing a point in a given time window