Check the answers to this question about letters

permutationsprobability

I just have question for checking if my answer is correct. Here is the question: Suppose four letters are chosen from "A, B, C, D, E" at random to form a word. The letters can be reused.

  1. What is the total number of outcomes in this experiment
  2. What is the probability the four letter word begins and ends with a vowel
  3. What is the probability that the word contains all distinct letters
  4. What is the probability that the four letter word begins and ends with a vowel and contains all distinct letters

My answer is:

  1. 6^4 =1296
  2. There's 2 vowels, A, E. It's arrange like that "A _ _ E" or "E _ _ A" or "A _ _ A" or "E _ _ E". Together is 6^2 times 4. So 144 times. The total is 1296 possibility so it's 11.11%
  3. This is 6 permute 4, 360. So it's 27.77%
  4. This is "A _ _ E" or "E _ _ A", and inside it is 4 permute 2, 24 possibility. And it's 1.85%

I want to ask if everything is right? Thanks.

Best Answer

I will express all of the probabilities as

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$

where, for all of the probability problems, I will use $D = 5^4.$

This means that $N$ must be computed in a consistent manner. So, since the denominator is distinguishing between (for example) "A-B-C-D" and "B-A-C-D" the numerator must make the same distinction.

(1) $5^4$.
There are $5$ choices for each of the $4$ letters.

(2) $N = (2^2) \times (5^2)$.
There are $2$ choices for each of the outer letters
and $5$ choices for each of the inner letters.

(3) $N = 5 \times 4 \times 3 \times 2 = 5!$.
There are $5$ choices for the 1st letter, then $4$ choices for the 2nd letter, and so on.

(4) $N = (2 \times 1) \times (3 \times 2).$
There are $2$ choices for the 1st letter,
then $1$ choice for the last letter.
Then, there are $3$ choices for the 2nd letter,
then $2$ choices for the 3rd letter.

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