7 October
2019
A quick puzzle
Two students independently come to the library every day after dinner between 7pm and 9pm. Each stays for exactly one hour and then leaves. At any night, what is the probability that they met? (meaning that both were in the library at the same time even if for a very short while)
One?
No. Think about it this way. First student comes at 7 and leaves at 8. Second student comes at 8:30 and leaves at 9:30. They never meet. So cannot be one.
3/4. The probably of first arrival between 7 and 8 is 1/2 and for them to meet the second arrival should happen within the next one hour is 1/2 again so this makes it 1/4. If the first arrival happens between 8 and 9 they always meet and the probability of the first arrival between 8 and 9 is again 1/2. Total probability is 1/2 + 1/4 which equals 3/4.
I was going to ask if they could stay beyond 9. Based on the answer by Sanjeev Mehta above, guessing they could.
If it was to be contained within 9 pm then it’s 100% right?
Each one stays 1 hr out of a 3 hr window, the probability of each one being there at any given moment (7-10) is 1/3 rd. So independent events both happening together is 1/9 th.. (I am too old to take these kind of reputational risks in the privacy of facebook, but if it entertains Rajib on a Monday..I will ).
Solution:
Here is a graphical way to think about the problem. Let’s plot on X-axis when the first person arrives and on Y-axis when the second person arrives. So the area in between is all the possible combinations of the two persons’ arrivals. Now, plot which part of the area they actually get to meet. If the first person comes at 7 (and leaves at 8) then as long as the first person comes between 7 and 8 (consequently leaves between 8 and 9), they meet. Similarly if first person comes at 8 (and leaves at 9), then as long as the second person comes between 7 and 9 (consequently leaves between 8 and 10), they meet. If you plot it, the area will look like the black shaded portion. The highlighted portion is when they DO NOT meet – which from geometry you can see is one fourth of the area. So, the probability they meet is 3/4
Solution:
Here is a graphical way to think about the problem. Let’s plot on X-axis when the first person arrives and on Y-axis when the second person arrives. So the area in between is all the possible combinations of the two persons’ arrivals. Now, plot which part of the area they actually get to meet. If the first person comes at 7 (and leaves at 8) then as long as the first person comes between 7 and 8 (consequently leaves between 8 and 9), they meet. Similarly if first person comes at 8 (and leaves at 9), then as long as the second person comes between 7 and 9 (consequently leaves between 8 and 10), they meet. If you plot it, the area will look like the black shaded portion. The highlighted portion is when they DO NOT meet – which from geometry you can see is one fourth of the area. So, the probability they meet is 3/4