# Calculate conditional probability with exponential distribution

Two computer specialists are completing work orders. The first specialist receives 60% of all orders.Each order takes her exponential amount of time with parameter λ1 = 3 hrs^-1. The second specialist receives the remaining 40% of orders. Each order takes him exponential amount of time with parameter λ1 = 2 hrs^-1.

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

Then I found an answer which included the lines:

A ={ first specialist processes the order }
B ={ second specialist processes the order }
C ={ the order takes more than 30 minutes (1/2 hr)}

Using Exponential distributions with parameters λ1 = 3 hrs^−1 and λ2 = 2 hrs^−1,
P{C | A}=e^(−(3)(1/2)) =e^(−1.5) and P{C | B}=e^(−(2)(1/2)) =e^(−1)


What I don't understand is how they came up with for example " P{C | A}=e−(3)(1/2)", it says nothing in the book I am reading that you can write it like that. The only thing I found in my book is that you could write:

P{C | A} = P{C ∩ A} / P{ A }

Exponential distribution: f(x) = λe^(-λ*x)


Can someone please explain how they ended up at the conclusion that P{C | A} = e^(−1.5) or point me in some direction where they explain it? Also another smaller question I have is why in questions is for example λ1 = 3 hrs^(−1) and not just λ1 = 3 hrs?

It seems that the notation $$\text{hrs}^{-1}$$ refers to the rate parameter, so the first specialist processes orders at a rate of $$3$$ per hour. The exponential distribution models time, and its mean is the reciprocal of the rate, so if the service rate is $$3$$ orders per hour, then the mean service time is $$1/3$$ hours per order.

Then, the quantity $$P\left(C|A\right)$$ is the probability that the order takes more than half an hour, given that specialist $$A$$ is processing it. For $$X\sim Exp\left(\lambda\right)$$, you have $$P\left(X > t\right) = e^{-\lambda t}$$. Letting $$X \sim Exp\left(3\right)$$, it follows that $$P\left(C|A\right)$$ is the same as $$P\left(X > 1/2\right) = e^{-3/2}$$. Likewise, $$P\left(C|B\right) = e^{-2/2} = e^{-1}$$. What you want, however, is

$$P\left(A|C\right) = \frac{P\left(A\cap C\right)}{P\left(C\right)} = \frac{P\left(C|A\right)P\left(A\right)}{P\left(C|A\right)P\left(A\right) + P\left(C|B\right)P\left(B\right)}$$

which is obtained by applying the definition of conditional probability repeatedly. All of these quantities can be straightforwardly computed now.

• But how do you arrive at "𝑃(𝑋>𝑡)=𝑒^(−𝜆𝑡) " when the formula for exponential distribution is f(x) = λe^(-λ*x)? Where does the λ go? Commented Oct 6, 2020 at 8:36
• The expression you give is the density. If you integrate it from $t$ to infinity, you arrive at $e^{-\lambda t}$. Commented Oct 6, 2020 at 12:59