Hi Kája, thank you for your comment. I’m glad you liked the post.

For the bus arrival example, the lambda is 4 (arrival in every 15 min = 4 events per hour).
If we plug lambda as 4:
P(T < 10 mins) = P(X=0 in 1/6 time units) = 1- e^−(4*0.166) = 1 -e^-0.6667 = 1- 0.5134 = 0.4865

Here is the relevant part from the post.

P(T > t) = P(X=0 in t time units) = e^−λt* T : the random variable of our interest!
the random variable for the waiting time until the first event
* X : the # of events in the future which follows the Poisson dist.* P(T > t) : The probability that the waiting time until the first event is greater than t time units
* P(X = 0 in t time units) : The probability of zero successes in t time units

For the second question: 2. Ninety percent of the buses arrive within how many minutes of the previous bus?

You can solve for t from 1-e^(-4*t)=0.9

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