Suppose a shop has two cashiers and one line of customers waiting to be served. The time for each cashier to serve the customer is an exponential random variable with \lambda = 1/4 and afterwards a customer leaves the store. Suppose both cashiers are busy with customers A and B, and customer C is the first in line to be served. What is the probability that among A, B, and C, C will not be the last to leave the store? Explain carefully. This question does not require extensive calculations.

\frac{17}{24}

Step-by-step explanation:

Given that Box 1 contains 2 red balls and 1 blue ball. Box 2 contains 3 blue balls and 1 red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn.

P(selecting Box 1) = 0.5 = P(selecting II box) (assuming a fair coin is tossed)

P(Red/Box I ) =

P(Red/Box II) =

Box 1 and Box 2 are mutually exclusive and exhaustive events.

So probability of selecting a red ball.

=  probability of selecting a red ball from box 1 +  probability of selecting a red ball frm box 2