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Problem 91

Combinatorics Difficulty 1.2 Multiple choice CEMC Gauss (Grade 8) · Canada · 2013

Each letter of the English alphabet is written on a separate tile and placed in a bag. Alonso draws one letter at random from the bag. What is the probability that Alonso draws a letter that is in his name?

Pick one

Official solution

Solution 1

To determine the smallest number in the set $\left\{ \frac{1}{2}, \frac{2}{3}, \frac{1}{4}, \frac{5}{6}, \frac{7}{12} \right\}$ , we express each number with a

common denominator of 12. The set $\left\{ \frac{1}{2}, \frac{2}{3}, \frac{1}{4}, \frac{5}{6}, \frac{7}{12} \right\}$ is equivalent to the set $\left\{ \frac{1\times6}{2\times6}, \frac{2\times4}{3\times4}, \frac{1\times3}{4\times3}, \frac{5\times2}{6\times2}, \frac{7}{12} \right\}$ or to the set $\left\{ \frac{6}{12}, \frac{8}{12}, \frac{3}{12}, \frac{10}{12}, \frac{7}{12} \right\}$ .
The smallest number in this set is $\frac{3}{12}$ , so $\frac{1}{4}$ is the smallest number in the original set.

Solution 2

With the exception of $\frac{1}{4}$ , each number in the set is greater than or equal to $\frac{1}{2}$ .

We can see this by recognizing that the numerator of each fraction is greater than or equal to one half of its denominator.
Thus, $\frac{1}{4}$ is the only number in the list that is less than $\frac{1}{2}$ and so it must be the smallest number in the set.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.