A young man and woman plan to meet between 5:00 and 6:00PM, each agreeing not to wait more than 10 minutes for the other person. What is the probability that they will meet if they arrive independently at random times between 5:00 and 6:00?
2006-09-21 16:59:39 · 9 answers · asked by Math_Guru 2 in Science & Mathematics ➔ Mathematics
assuming each arrive at the start of the minute. also, the last time to arrive is 5:50 so that he leaves at 6:00. There are 51 "minute marks" between 5:00 - 5:50.
what is the probability that A arrives
(a) exactly at 5:00 (wait until 5:09): 1/51
- what is the probability that B arrives between 5:00 - 5:09: 10/51
- prob = 1/51 X 10/51 = 10/2601
(b) exactly at 5:01 (wait until 5:10): 1/51
- what is the probability that B arrives between 5:00 - 5:10: 11/51
- prob = 1/51 X 11/51 = 11/2601
(c) exactly at 5:02 (wait until 5:11): 1/51
- what is the probability that B arrives between 5:00 - 5:11: 12/51
- prob = 1/51 X 11/51 = 12/2601
(d) 5:03 = 13/2601
(e) 5:04 = 14/2601
(f) 5:05 = 15/2601
(g) 5:06 = 16/2601
(h) 5:07 = 17/2601
(i) 5:08 = 18/2601
(j) 5:09 = 19/2601
(k) 5:10 = 19/2601
.
.
.
5:40 = 19/2601
5:41 = 19/2601
5:42 = 18/2601
5:43 = 17/2601
5:44 = 16/2601
5:45 = 15/2601
5:46 = 14/2601
5:47 = 13/2601
(x) 5:48 = 12/2601
(y) 5:49 = 11/2601
(z) 5:50 = 10/2601
total = 879/2601 ~ 33.8%
2006-09-21 18:00:56 · answer #1 · answered by Anonymous · 1⤊ 1⤋
11/36, or 0.30555_.
If person 1 arrives at any time within the middle 40 minutes of the hour, then they have 1/3 of a chance of meeting the other person (it scans the ten minutes on both sides), which comes to 2/9, or 8/36. If they arrive within 10 minutes of the end of the hour, then they average 15 minutes with which to mee the other person, rather than 20, so that's 1/3 of a chance of having 1/4 of a chance to meet them, which is 1/12, or 3/36. 3/36 and 8/36 makes 11/36 of a chance to meet the other person.
2006-09-21 17:06:55 · answer #2 · answered by Anonymous · 3⤊ 0⤋
1 in 6
2006-09-21 17:02:24 · answer #3 · answered by gone 7 · 0⤊ 4⤋
1 in 12
2006-09-21 17:18:03 · answer #4 · answered by Anonymous · 0⤊ 3⤋
12+13+7+5= 37 shirts in finished the threat that you first grab a pink one is: 12/37 blue one is: 13/37 eco-friendly one is: 7/37 orange one is: 5/37. besides, the threat that he does no longer word is 0, because no count number the colour of the shirts you grab, there's a threat more advantageous than 0 that he will word.
2016-11-23 14:23:20 · answer #5 · answered by turick 3 · 0⤊ 0⤋
Rather than a mathematical solution, I'll give a conceptual solution.
The first person to arrive will wait ten minutes - which is one-sixth of the total 1-hour window.
Thus, the second person has a 10-minute window (out of sixty) in which to arrive to make the connection.
Therefore, the chance of them meeting is 10 minutes out of 60 = one sixth = 16.6%
2006-09-21 17:06:23 · answer #6 · answered by Anonymous · 2⤊ 3⤋
50%; 1 in 2. Get a ruler and measure a line 6 inches long. Cut two pieces of paper each 1 inch long. Slide them along the line...
2006-09-21 17:27:47 · answer #7 · answered by zee_prime 6 · 0⤊ 3⤋
because this is a very simple question, the answer is 1 in 6
2006-09-21 17:15:31 · answer #8 · answered by soar_2307 7 · 0⤊ 3⤋
i think the ans. is 109/ 720 interesting problem
2006-09-21 17:30:18 · answer #9 · answered by the wise one 3 · 1⤊ 1⤋