Difference between revisions of "2009 Grade 8 CEMC Gauss Problems/Problem 9"

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-==Problem==
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-If Jeff picks one letter randomly from the alphabet, what is the probability that the letter is in the word "probability"?
-<math> \text{ (A) }\ \frac{9}{26} \qquad\text{ (B) }\ \frac{15}{26} \qquad\text{ (C) }\ \frac{10}{26} \qquad\text{ (D) }\ \frac{17}{26} \qquad\text{ (E) }\ \frac{8}{26} </math>
-==Solution==
-We can count the amount of letters in the word probability (excluding repeated ones, such as the second b), and divide the result by the total number of letters in the alphabet.
-In the word probability, p, r, o, a, l, t, and y all appear once, but b and i appear twice. This means that there are <math>9</math> unique letters in the word.
-In the alphabet, there are <math>26</math> letters. Thus, the probability of the letter being in the word, "probability", is <math>\boxed {\textbf {(A) } \frac{9}{26}}</math>.
-~anabel.disher