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 λ2 = 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?

0.6

Step-by-step explanation:

Data:

Let W be the event the order is not ready in 30 mins.

Then, let A be the event the first worker got the order. So, we want the probability, P(A/W). Using formula for conditional probability gives:

In this case, we need to calculate a couple of probabilities.

The event that W can happen can be like this:

1. the first worker got the order

2. the second worker got the order and it is not ready.

the probability for both events are disjoint, so:

for item (1) the probability will be 0.6 times the probability that the first worker takes 30 mins to do the job.

Calculating:

The probability that an exponential with parameter is 2 is given by:

so, the probability is

thus, the probability is > t is

That's the same as 0.6

28

step-by-step explanation:

8+8+6+6

answer: idk

step-by-step explanation: