Question 1169230: A permutation of the word "white" is chosen at random. Find the probability that it begins
with a vowel. Also find the probability that it ends with a consonant.
Answer by math_tutor2020(3817)   (Show Source):
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Part 1
There are 2 vowels (i and e), so we have 2 choices for the first slot. After picking a vowel, we have 4 letters left. Let's say we pick 'i' as the first letter. That means we have w,h,t, or e left over. A very similar situation happens with picking 'e' as the first letter.
So we have 2 choices for slot A, and 4 choices for slot B. So far we have 2*4 = 8 different permutations.
For slot C we'll have 3 letters left to pick from, then slot D gets 2 choices, finally slot E only gets one choice (whatever letter is left over). We have the values counting down by 1 each time we move to the next slot.
So we have 2*4*3*2*1 = 48 different permutations in which we have a vowel as the first letter.
This is out of 5*4*3*2*1 = 120 different permutations possible of rearranging five letters
The probability of getting a vowel as the first letter is 48/120 = 2/5
Final Answer: 2/5
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Part 2
Here's the set of consonants we have to work with: {w, h, t}
So we have three choices to select for slot E
Whatever is picked for slot E leaves...
4 choices for slot A
3 choices for slot B
2 choices for slot C
1 choices for slot D
Note we could easily rearrange the slot labels and say something like
4 choices for slot D
3 choices for slot C
2 choices for slot B
1 choices for slot A
and we get the same basic idea going on.
All that matters is that we're multiplying out the correct values, which is 3*4*3*2*1 = 72
We have 72 different ways to have a consonant at the end out of 120 different permutations of five letters.
So, 72/120 = 3/5 is the probability of getting a consonant as the last letter.
Final Answer: 3/5
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