Prof. Stoiciu’s math puzzle night is held every Tuesday between 6 and 7 PM in the Dennett private dining hall in Mission. All students but one arrives to the event at 6PM. The missing student arrives sometimes between 6 and 7 PM. Prof. Stoiciu wants to give an opportunity for this student to solve problems but at the same time wants to be fair with others, so he looks at the clock and randomly picks a minute between 6 and 7PM. In this minute he gives out the weekly problem-sets to the students present. Prof. Stoiciu and the initial students (excluding the missing one) are happy when the missing student arrives no sooner than 15 minutes before the randomly selected minute. In this way they don’t feel like “wasting so much time even if the missing one is present.” The missing student is happy when he/she arrives no later than 15 minutes after the randomly selected minute, since he/she can still get the problem-set and solve couple problems. What is the probability that Prof. Stoiciu and all students (including the missing one) will be happy?

Thank you very much for sending in your solutions and congratulations to Faraz Rahman for submitting the correct solution with proper explanation.

Faraz’s Solution:

Since we’re dealing with a one hour window of time, from 6pm to 7pm, we can scale the times picked by Prof. Stoiciu and the student such that, for example, 6pm becomes 0, 6:45pm become 0.75, 7pmbecomes 1, and so on. For any time picked by Prof. Stoiciu, x, our conditions will be satisfied if the time, y, at which the student arrives is between x-0.25 and x+0.25. If we were to graph these two lines, y=x-0.25 and y=x+0.25, in our 1×1 “box”, then the area enclosed between the two lines in the box will give us the required probability. We see that the area within the box which is outside that enclosed by the two lines is made up of two triangles. Each of these has area (1/2)(3/4)(3/4), so together they have a combined area of 9/16. Deducting this from 1 gives us 1-9/16 = 0.4375, which is the required probability.

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