Alice, Bob, and Carl arrange to meet for lunch on a certain day. They arriveindependently at uniformly distributed times between 1 pm and 1:30 pm on that day. Required:
a. What is the probability that Carl arrives first?
b. What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?
c. What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?
d. What is the probability that the person who arrives second will have to wait more than 5 minutes for the third person to show up?

a. What is the probability that Carl arrives first?

Probability that Carl arrives first is ¹/₃ = 33.33% since their arrival times is uniformly distributed. The same probability applies to Bob and Alice.

b. What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?

Assuming that Carl arrived on time, 1:10 PM, we must determine the probability that Alice or Bob arrive between 1:20 and 1:30 (half the remaining time)

P = [3 · (¹/₂ - ¹/₃)] · [3 · (¹/₂ - ¹/₃)] = (3 · ¹/₆) · (3 · ¹/₆) = ¹/₂ · ¹/₂ = ¹/₄ = 25% chance that either Alice or Bob arrive more than 10 minutes later

c. What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?

P = 1 - 25% = 75%

d. What is the probability that the person who arrives second will have to wait more than 5 minutes for the third person to show up?

I divided the 20 minutes by 5 to get ¹/₄:

P (|S - T| ≤ ¹/₄) = {[(x + ¹/₄)²] / 2} + (1 / 2x) + {[(⁵/₄ - x)²] / 2} = 0.09375 + 0.25 + 0.09375 = 0.4375 = 43.75%

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3