Naughty Willy

Probability Level 4

One day, at precisely 9:30 AM, Willy loses Arman's Student ID card. It turns out that the ID was lost in Dr. Wu's Algebra class. Dr. Wu, confused, puts the ID on the board and draws an arrow saying "Arman's ID".

Willy knows he has made a tremendous mistake, so he immediately tells Arman. Enraged, Arman gives Willy a warning and then searches the school for his ID.

It is given that Arman will go to Dr. Wu's classroom at a random time between 10 AM and 11 AM and see his ID. However, Dr. Wu will also check his board at a random time between 10 AM and 11 AM. If he sees that the ID is not there, he will not worry about it, but if he sees the ID still there, he will wait 15 more minutes for Arman to come and then throw it away.

The probability that Arman's ID is thrown away can be expressed as a common fraction a b \frac{a}{b} . Find the product a b ab .


The answer is 288.

William Cui by William Cui , 21, USA

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1 solution

Edward Rong
Jun 13, 2014

Let the minute during the 10:00 to 11:00 period at which Arman goes to collect the ID be x x . For example, x = 30 x=30 would correspond to 10:30. It should be noted that x x can be any real number between 0 and 60.

Let the minute during the 10:00 to 11:00 period at which Dr Wu randomly checks the board be y y . Similary, y can be any real number between 0 and 60.

In order for Arman's ID to be thrown away, it must be that the time at which Arman goes to collect the ID is more than 15 minutes after Dr Wu checks the board (since he throws away the ID 15 minutes after looking at the board) i.e. x > y + 15 x > y+15

To help us visualise this, let us draw a graph within the range 0 ≤ x ≤ 60 0 \leq x \leq 60 and 0 ≤ y ≤ 60 0 \leq y \leq 60 . On this graph, we plot the equation x = y + 15 x = y+15 and the area that corresponds to the inequality x > y + 15 x > y+15 is the area under this line. This area is a right-angled isoceles triangle with base and height both 45 45 units, and therefore the area is 45 × 45 2 \large\frac{45\times45}{2}

The total number of possibilities for the times at which Arman goes to collect the ID and Dr Wu looks at the ID is 60 × 60 60\times60

Therefore, the probability that Arman's ID is thrown away is: 45 × 45 2 60 × 60 = 3 × 3 2 4 × 4 = 9 32 \LARGE \frac{\frac{45\times45}{2}}{60\times 60}=\frac{\frac{3\times3}{2}}{4\times4}=\frac{9}{32}

Therefore, a b = 9 × 32 = 2 88 ab=9\times32=\mbox 288

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