A local Starbucks coffee shop has two baristas filling orders, and it is known that the time it takes for each barista to complete an order is exponentially distributed with average 60 seconds. You are next in line with barista 1 filling an order for John and barista 2 filling an order for Mary. When one of these two customers gets their order, you will proceed to that barista to get your order filled. What is the probability that, of these three customers, you will be the last one of the three to leave the coffee shop with your coffee order

We know that if X ~ Exp(\lambda_x) and Y ~ Exp(\lambda_y), then

P(X < Y) = \lambda_x / (\lambda_x + \lambda_y )

Let T1 and T2 be the time taken for barista 1 and barista 2 filling an order respectively.

T1 ~ Exp(1 per 60 seconds) = T1 ~ Exp(1 per minute)

Similarly, T2 ~ Exp(1 per minute)

Then,

P(T1 < T2) = 1 / (1 + 1) = 1/2

P(T2 < T1) = 1 / (1 + 1) = 1/2

Probability that you will be the last one of the three to leave the coffee shop with your coffee order

= Probability that John (barista 1) leaves before Mary (barista 2) and Probability that Mary (barista 2) leaves before you (barista 1) + Probability that Mary (barista 2) leaves before John (barista 1) and Probability that John (barista 21) leaves before you (barista 2)

= P(T1 < T2) * P(T2 < T1) + P(T2 < T1) * P(T1 < T2) (Due to Memoryless property of exponential distribution)

= (1/2) * (1/2) + (1/2) * (1/2)

= 1/2

Step-by-step explanation:

c

step-by-step explanation:

c

my dear correct answer is option number 3