Answer :
Answer:
The probability that the first to arrive waits no longer than 15 minutes is 1/3
Step-by-step explanation:
Let X be the distribution of time where the man arrives and Y the distribution for the woman. Since it is a 40 minute period, the density function of X is
[tex]f_X(t) = 1/40 \, I_{[11:30, 12:10]}[/tex]
(Note: it is weird not to work with real numbers, however, you can think an hour as 60 minutes; for example 11:30 = 11*60+30 = 690, so we are working with real numbers after all. But it is more comfortable to work with minutes and hour notation)
The density of Y is
[tex]f_Y(t) = 1/45 \, I_{[11:50, 12:35]}[/tex]
Note that if the man arrives at 11:30, he will have to wait more than 15 minutes regardless of when the woman arrives; he has to arrive at least 11:35 for a chance. The same way, no matter how late the man arrives, he will wait more than 15 minutes if the woman arrives after 12:25
We can obtain the desired probability spltting the event in 2: when the man arrives earlier and when the woman does. For the first even we need to integrate the product of the density functions (which are independent), with X taking any value greater than 11:35 until 12:10 and Y being between X and X+15. Then we integrate with X between 11:50 + 15 = 12:05 and 12:10 and Y between X-15 and X
[tex]P(|X-Y| \leq 15) = P(0 \leq X-Y \leq 15) + P(0 \leq Y-X \leq 15)[/tex]
Now,
[tex]P(0 \leq X-Y \leq 15) = \int\limits^{12:10}_{12:05}\int\limits^{x}_{x-15} {\frac{1}{45*40}} \, dy \, dx = \int\limits^{12:10}_{12:05} \frac{x - (x-15)}{1800} \, dx = \\\frac{(12*60+10) - (12*60+5)}{120} = 5/120 = 1/24[/tex]
And,
[tex]P(0 \leq Y-X \leq 15) = \int\limits^{12:10}_{11:35}\int\limits^{x+15}_{x} {\frac{1}{1800}} \, dy \, dx = \int\limits^{12:10}_{11:35} \frac{(x+15) - x}{1800} \, dx = \\\frac{(12*60+10) - (11*60+35)}{120} = 35/120 = 7/24[/tex]
Thus, P(|X-Y| ≤ 15) = 1/24 + 7/24 = 1/3.
The probability that the first to arrive waits no longer than 15 minutes is 1/3.
Answer:
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