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Find the probability that a random arrangement of QUOTHTHERAVENNEVERMORE has:

a) The first V before the first R and the vowels in alphabetical order.

b) At least 2 other letters in between successive E's.

I'm quite confused and I could use some help on how to make sense of these problems (a and b are separate problems).

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    Don't start your question in the title, it's unnecessary and makes it hard to parse. Add a simple, descriptive title and write a self contained question. See the help guidelines – 2017-02-10
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    If I am counting correctly (not sure) there are two $V's$ and three $R's$, so the probability that the first one of those you draw is a $V$ is $\frac 25$. – 2017-02-10
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    Are $a)$ and $b)$ two separate problems? – 2017-02-10
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    Yes, a and b are not related. – 2017-02-10

1 Answers 1

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For a), we can consider the events on the consonants and the vowels independently. You have one $A$, five $E$, two $O$, and one $U$. This is a total of nine vowels. There are $9!$ permutations of them, but removing symmetry we have only $\frac{9!}{2!5!}$ distinct ways to arrange them. The only alphabetical sequence on vowels looks like $AEEEEEOOU$, so the probability that a random vowel sequence is alphabetical is $\frac{2!5!}{9!}$.
For the consonants, we have two $V$ and three $R$. The first $V$ comes before the first $R$ iff $V$ is the first letter of the consonant sequence. There are $\frac{5!}{2!3!}$ sequences of consonants. If we pick $V$ as the first letter, there are $\frac{4!}{3!}$ sequences on the next $4$ letters (1 $V$, 3 $R$). Dividing we see that $\frac{2}{5}$ of the consonant sequences begin with $V$.

Since the probabilities of the two conditions in a) are independent, the probability of both conditions being satisfied is the product of their individual probabilities. This is $4\frac{4!}{9!}$