# Question #e7870

Jan 12, 2018

$\frac{37}{42}$

#### Explanation:

Let's find the opposite outcome: what's the probability that no vowels are chosen?

Well, BREAKDOWN has 9 letters, and 6 are consonants. The probability that the first letter chosen being a consonant is $6 / 9$.

Now assume we choose a second letter. We have left only 8 letters now, and since we already chose a consonant, only 5 consonants are left. The probability of picking a second consonant is $5 / 8$.

Using the same logic, the probability of choosing a third consonant is $4 / 7$ and a fourth consonant is $3 / 6$.

So, the probability of choosing all 4 letters that are all consonants is:

$\frac{6}{9} \left(\frac{5}{8}\right) \left(\frac{4}{7}\right) \left(\frac{3}{6}\right)$

Simplifying...

$= \frac{2}{3} \left(\frac{5}{8}\right) \left(\frac{4}{7}\right) \left(\frac{1}{2}\right) = \frac{5}{42}$

What we want, however, is the case where at least $1$ vowel is selected. This has zero overlap with the case we found, where $0$ vowels are selected.

Thus, the probability we are looking for is:

$1 - \frac{5}{42} = \frac{37}{42}$