# Geometric Sequence with Normal Distribution Problem

Given: The running time (in seconds) of an algorithm on a data set is approximately normally distributed with mean 3 and variance 0.25.

a. What is the probability that the running time of a run selected at random will exceed 2.6 seconds?

Answer for (a): I've computed this and found it to be p = 0.7881.

b. What is the probability that the running time of exactly one of four randomly selected runs will exceed 2.6 seconds?

Answer for (b): Not sure, I know it's a geometric sequence and I believe the formula that I need to use to be $p(1-p)^3$, thus giving me $0.7881(1-0.7881)^3$ however this was marked incorrect. I'm trying to figure out why, can someone explain my error and how to arrive at the correct solution?

Thanks!

-
You have 4 choices for which run will exceed 2.6 seconds. So your answer should be $4p(1-p)^3$ instead of $p(1-p)^3$. You basically calculated the probability that a pre-determined run will exceed 2.6 seconds, and the other 3 runs are less than 2.6 seconds.