# How to find probability based on percent of orders recieved and exponential amount of time to complete per order?

Two computer scientist are completing work orders. The first specialist receives 60% of all the orders. Each order takes her Exponential amount of time with parameter λ=3 hrs^-1. The second specialist receives the remaining 40% of orders. Each order takes him Exponential amount of time with parameters λ= 2 hrs^-1.

A certain order was submitted 30 minutes ago, and it is still not ready. What is the probability the first specialist is working on it?

I'm not fully sure how to tackle this problem but right now I am fooling around with Poisson and Exponential Distribution.

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Are you assuming that the specialists were free when the order arrived and started work on it immediately? – joriki Nov 6 '12 at 21:32
I'd assume that as soon as the order arrives the specialists works nonstop until completion. – jrquick Nov 6 '12 at 21:41

Let $W$ be the event the order is not ready in $30$ minutes. Let $A$ be the event the first worker got the order. We want $\Pr(A|W)$. By the usual formula for conditional probabilities, we have $$\Pr(A|W)=\frac{\Pr(A\cap W)}{\Pr(W)}.$$

So we need to compute a couple of probabilities. Let's tackle the harder one. The event $W$ can happen in two ways: (i) the first worker got the order, and it is not ready or (ii) the second worker got the order, and it is not ready.

These two events are disjoint, so we need to compute the probability of each and add. For the probability of (i), this is $(0.6)$ times the probability that the first worker takes more than $1/2$ hour to do a job. You can compute this probability by using the fact that the waiting time for the first worker has exponential distribution with parameter $3$.

The probability that an exponential with parameter $3$ is greater than $\frac{1}{2}$ is $e^{-3/2}$. So the probability of (i) is $(0.6)e^{-3/2}$. One can do this by integration, but you probably already know that if $t\ge 0$, then the probability that an exponential with parameter $\lambda$ is $\le t$ is $1-e^{-\lambda t}$, so the probability it is $\gt t$ is $e^{-\lambda t}$.

Calculating the probability of (ii) is very similar.

Finally, note that $\Pr(A\cap W)$ is just the probability of (i), so we have already found it.

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