Question

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?

Answer 1

Answer:

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:

$P (A/W) = \frac{P(AnW)}{P(W)}$

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:

$P = \frac{1}{2}e^-{\frac{2}2} }$

so, the probability is $0.6e^{-1}$

thus, the probability is > t is $e^{-\lambda t }$

That's the same as 0.6

Answer 2

The probability that the first specialist is working on it is 0.4764.

How to calculate the probability?

P(not ready in 30 minutes(0.5 hrs) will be:

=P(specialist 1) × P(not ready in 30 minutes |specialist 1) + P(specialist 2) × P(not ready in 30 minutes specialist 2)

= 0.6 × (e-0.5*3) + 0.4 × (e-0.5*2)

= 0.6 × 0.2231 + 0.4 × 0.3679

= 0.2810

Therefore P(first specialist given not ready in 30 minutes(0.5 hrs))

=P(specialist 1) × P(not ready in 30 minutes |specialist 1)/P(not ready in 30 minutes(0.5 hrs))

=0.6 × (e-0.5*3)/0.2810

=0.6 × 0.2231/0.2810

= 0.4764

Learn more about probability on:

https://brainly.com/question/24756209