as a noob when it comes to probability, i shall take a stab at this one.

we have independent events here, and we have two outcomes which are the same for each passenger involved (the bus stops for you or not). we can deem the bus stopping as a "success" and the bus not stopping as a "failure" and use the binomial distribution.

recall that the probability of successes in independent trials is given by:

where is the probability of success, and is the probability of failure

Now, you want , where is a discrete random variable for the number of stops the bus will make. so you add all the probabilities from 1 up to k inclusive.

Now recall that for a random variable , with probability mass function given by (here, that is given by the formula for the binomial distribution), has expected value:

now replace with the formula for the binomial distribution, and sum from goes from up to , where .