Question

Alice and Bob arrange to meet for lunch on a certain day at noon. However,neither is known for punctuality. They both arrive independently at uniformly distributedtimes between noon and 1 pm on that day. Each is willing to wait up to 15 minutes forthe other to show up. What is the probability they will meet for lunch that day? "slader"

Answer 1

The probability that Alice and Bob will meet for lunch is 2025/7200, or approximately 0.28125.

Step-by-step explanation:

To calculate the probability that Alice and Bob will meet for lunch, we can consider the possible scenarios. Let's assume that Alice arrives at time x minutes after noon, and Bob arrives at time y minutes after noon. The probability that they will meet for lunch is the probability that the absolute difference between x and y is less than or equal to 15 minutes.

If x is the larger of the two time values, then the probability that they will meet is given by the area of the trapezoid between the lines y = x - 15 and y = x + 15, within the square bounded by x = 0, x = 60, y = 0, and y = 60. This area is equal to (60-15)(60-15)/2 = 2025/2.

The same calculation can be done if y is the larger of the two time values, resulting in the same probability of 2025/2.

Therefore, the probability that Alice and Bob will meet for lunch is 2025/2 divided by the total area of the square, which is 60*60 = 3600. This simplifies to a probability of 2025/7200, or approximately 0.28125.

Answer 2

probability of there meeting = 15/(60-15) = 15/45 = 0.333 = 33%

Step-by-step explanation:

favorable event = +- 15 minutes

total time they can eat launch is between noon and 1pm = 60minutes

they are both known for non punctuality, that means they cant meet up at 12noon, let assume 15minutes pass noon.

probability of there meeting = 15/(60-15) = 15/45 = 0.333 = 33%

which is probably saying they will not meet for lunch that day.