Question

Two students agree to meet at a restaurant between 6pm and 7pm.  Suppose they each agree to wait 15 minutes or until 7pm (whichever comes first) for the other person before leaving.  What is the probability that they will meet if they each arrive independently and at random (uniformly) between 6pm and 7pm?  Round your answer to two decimal places.

Answer 1

p = 43.75%.

Explanation:

Suppose that student 1 arrives before the other.

The probability that student 1 arrives during the first 3/4 hour is 0.75. Then, he will wait 1/4 hour.

The probability that he arrives during the last 1/4 hour is 0.25, and then (on average he will wait) 1/8 hour. So altogether the student 1 will wait

(0.75)*(1/4) + (0.25)(1/8) = 0.21875.

Then, the probability that student 2 arrives while the student 1 is waiting is 0.21875.

In the same way, if student 2 arrives before student 1.

The probability is

p = (0.21875) + (0.21875) = 0.4375

⇒ p = 43.75%.

Answer 2

Answer:

7/16

Approximately 0.44

Step by step explanation:

For the 1st student to arrive before 2nd student,

* the probability that he(1st student)to arrive during 3/4 hrs = 3/4

*Then he(1st student) will need to wait for 1/4 hrs

*The probability that he(1st student) arrive during the last 1/4 hrs = 1/4

*Then he(1st student) waits for average of 1/8 hrs (1/4 + 1/4)=1/8

*The total time he(1st student) wait = (3/4 ×1/4) +(1/4× 1/8) = 7/32

*For 1st and 2nd student to meet, the 2nd student need to arrive when 1st student is still arround, therefore the probability that 2nd student arrives when 1st student is arround = 7/32

*If 2nd student arrives before 1st student, the probability that they would meet is 7/32

* Adding both together 7/32 + 7/32 = 7/16

ANSWER = 0.4375

=0.44