Question #3767

Two persons agreed to meet at a certain place between 2 PM and 3 PM. Each person has to wait another one for 10 minutes and then leave. What is the probability of their meeting if everyone can arrive whenever during this hour?

Expert's answer

I will use geometrical probability to solve this problem.

Consider a plane with two coordinate axes.

Let the value on first coordinate axis is time where the first person came, and on the second – when the second person came. Both of them will come inside 2 and 3 PM.

The time when they meet each other, is, obviously, the set of points (x,y) on plane, that satisfy |x-y|<1/6 (1/6 of hour is 10 minutes). This set of points is plotted below:

To find the probability we need to find ratio of area of red segment and area of the square.

Area of the square is equal to 1.

Area of red segment is equal to (1-2*S), where S is area of upper left triangle (or, equally, of right down triangle).

S=1/2*5/6*5/6=25/72 (because side of triangle is equal to 5/6).

Finally,

P(meeting)=1-2S=1-25/36=11/36.

Consider a plane with two coordinate axes.

Let the value on first coordinate axis is time where the first person came, and on the second – when the second person came. Both of them will come inside 2 and 3 PM.

The time when they meet each other, is, obviously, the set of points (x,y) on plane, that satisfy |x-y|<1/6 (1/6 of hour is 10 minutes). This set of points is plotted below:

To find the probability we need to find ratio of area of red segment and area of the square.

Area of the square is equal to 1.

Area of red segment is equal to (1-2*S), where S is area of upper left triangle (or, equally, of right down triangle).

S=1/2*5/6*5/6=25/72 (because side of triangle is equal to 5/6).

Finally,

P(meeting)=1-2S=1-25/36=11/36.

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