I'm having trouble with the questions below, all relating to string probability. I'll write the problem and then provide my work for my (incorrect) answer. Please help me figure out what I did wrong.

SET1: Assume a string, allowing repetition, is randomly selected from all strings of length 4 from the set A = {r, e, a, s, o, n}.

Q1. What is the probability that it contains exactly one letter that is a vowel?

A1: 6^4 total probability, thus one vowel should be equal to... (C(4,1) * C(4,3)) / (6^4)

Q2. What is the probability that such a string contains exactly two r's given that it contains exactly two o's?

A2: two r's and two o's in a string of length 4 have 6 permutations, so according to conditional probability, we must show P(two r's | two o's) = P(two r's and two o's) / P(two o's), thus we have... (6 / (6^4)) / (C(4,2) * ((1/6)^2) * C(4,2) * ((5/6)^2))

SET2: How many distinct permutations of the letters in "letters"...

Q1. Begin with two vowels?

A1: 2 * 1 * 5 * 4 * 3 * 2 * 1 = 2 * (5!), since there are two vowel choices for the first spot, one for the second spot, then five remaining letters, then four, etc...

Q2. Begin with two e's or end with two t's?

A2: 2(2 * (5!)) - (2 * 1 * 3 * 2 * 1 * 2 * 1), used similar logic to the logic explained above in A1.

Q3. Have the vowels together?

A3: 6!, since you group the vowels together as one entity then find a place for all 6 entities.

Thank you!

p.s. If someone could format this, I'm not familiar with latex formatting and didn't see anything about math formatting in the advanced formatting help link. :-/

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    $\begingroup$ For the first, the vowel can be chosen in $3$ ways, its location in $4$ ways, and the remaining slots can be filled with consonants in $3^3$ ways, so the probability is $(3)(4)(3^3)/6^4$. For the second, your expression is hard to read, one should get $1/25$. $\endgroup$ – André Nicolas Mar 13 '14 at 22:43
  • $\begingroup$ Thanks for the effort on the first one, could you explain it using combinations as I did so I can understand exactly where I went wrong? That would be great. An answer alone for the second doesn't do me much good though since I got it wrong and don't know how to find the 1/25 probability. :( I'm really looking to learn from my mistakes. $\endgroup$ – CODe Mar 13 '14 at 22:57
  • $\begingroup$ My solution used combinations, the vowel can be chosen in $\binom{3}{1}$ ways, and its location can be chosen in $\binom{4}{1}$ ways. For every way of settling the kind and location of the vowel, there are $3^3$ permutations of the consonants that make up our $4$-letter word. $\endgroup$ – André Nicolas Mar 13 '14 at 23:03
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    $\begingroup$ If you are going to use the conditional probability formula for the second, let's do probability of $2$ o's first. The $2$ o's can be place in $\binom{4}{2}$ ways. The remaining $2$ slots can be filled with non-o's in $5^2$ ways, so the probability is $\binom{4}{2}5^2/6^4$. For the probability of $2$ o's and $2$ r's, the only freedom we have is where the o's go, this can be chosen in $\binom{4}{2}$ ways, the probability of $2$ of each is $\binom{4}{2}/6^4$. $\endgroup$ – André Nicolas Mar 13 '14 at 23:08

For problem Set1:

I will use $v$ for vowel and $c$ for consonant. For (1), you have strings of length $4$. So $s = _ _ _ _$. You want a single vowel. Suppose $s = v _ _ _$. The remaining three spots are consonants. There are three consonants, with one consonant per slot. So $s = v c c c$. So the number of ways to get $v = \binom{3}{1} = 3$, and the number of ways to get each $c$ is also $3$. So there are $3 * 3^{3} = 3^{4}$ ways to form a string with the first letter a vowel.

Now consider if $v$ is in the second position. The number of such strings is the same as if $v$ was in the first position. So you can easily see that the answer to your question is $3^{4} * 4$.

For the problem Set2:

For your first problem with "letters", there are two e characters. So by the multinomial counting rule, you divide by 2. Thus, the number of ways to start with two vowels is 5!.

You are making the same mistake in the second problem. Since there are two t's, you have to divide by 2. Thus, you have 2∗5!−3! as your answer. Your thinking is correct, though. Count first the number of ways to get two e's upfront (which is the answer to your first problem), then count the number of ways to get two t's at the end (also the answer to your first problem), then subtract out the number of ways to start with two e's and end with two t's (ie., permute the middle 3 characters).

Your answer for part 3 is correct.

  • $\begingroup$ This definitely helps clarify for Q1 for the first bolded assumption. Thanks very much for taking the time to explain fully. If you could help me with the others, I'd greatly appreciate it! :) $\endgroup$ – CODe Mar 13 '14 at 23:36
  • $\begingroup$ I've compiled your two separate answers together, please accept the edit and I'll accept this as the full answer. Thanks very much for the help! $\endgroup$ – CODe Mar 14 '14 at 0:19
  • $\begingroup$ I'm happy with the suggested edit, but it seems I cannot accept it. Glad you found my posts useful! $\endgroup$ – ml0105 Mar 14 '14 at 0:45

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