Problem of the Day

If the letters in the word ‘cucumber’ are rearranged in a random order, what is the probability that the new sequence of letters will begin and end with a vowel?

A. \frac{1}{4}

B. \frac{3}{8}

C. \frac{3}{28}

D. \frac{45}{56}

E. \frac{7}{8}

Reveal Answer



C. $\frac{3}{28}$

See the Solution



Of the 8 letters in cucumber, 3 of them are vowels: 2 u’s and 1 e. ¬†Therefore, a randomized sequence of these letters would begin¬†with a vowel $\frac{3}{8}$ of the time. ¬†Of those times, there would be two remaining vowels to distribute among seven remaining spots. ¬†So, the probability of the last spot being filled with a vowel would be $\frac{2}{7}$. ¬†Multiplying these two probabilities together gives us the answer we are looking for: $\frac{3}{8} \cdot \frac{2}{7}=\frac{6}{56}=\frac{3}{28}$.

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