A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:30 and 12:10 and if the woman independently arrives at a time uniformly distributed between 11:50 and 12:35, what is the probability that the first to arrive waits no longer than 15 minutes?

5/21

Step-by-step explanation:

From the question and time interval given the Man will most likely arrive before the woman so i will breakdown the various possible arrival time of the Man in relation to the woman's arrival time.

lets say

11:40 to 11:50

probability of arrival by the man = 10/35

during this period the woman has a probability of arriving less than 10 minutes within a 5 minutes interval

her own probability = 5/60

11:50 to 12:00

probability of arrival by the man = 10/35

during this period the woman has a probability of arriving within 10 minutes of the man's arrival within a 15 minute interval

her probability of arrival = 15/60

12:00 to 12:15

probability of arrival by the man = 15/35

during this period the woman has a probability of arriving within 20 minutes interval

her probability = 20/60

therefore to get the probability that the first to arrives waits no longer than 15 minutes

we multiply the various probabilities for each time interval and then add them up

= ( 10/35 * 5/60 ) + ( 10/35 * 15/60 ) + ( 15/35 * 20/60 )

= 500/2100

= 5/21

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3