enter image description here For the following question I figured that the expected time between successive arrival is the mean = 1/10 per hour (or 1 per 6 minutes). My question is regarding the second part; does the fact that the last driver has arrived 15 minutes ago effect any calculations? if not will the answer simply be P(x>0.25)=1-P(x<=0.25)=0.024? (P is Cumulative Distribution Function)

If the information given regarding the last drive-up is relevant could you provide me with a hint?



The random variable $X$ has exponential distribution with parameter $10$, or equivalently mean $\frac{1}{10}$.

The exponential distribution is memoryless. The fact that we have waited $\frac{1}{4}$ (of an hour) has no effect on the distribution of the additional waiting time.

Note that we want the probability that an exponential with parameter $10$ is $\gt \frac{1}{4}$. This is $$\int_{1/4}^\infty 10e^{-10 t}\,dt,$$ which is $e^{-2.5}$. (That is not the number you obtained.)


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