Question 1127738: if a permutation of the word "white" is selected at random, find the probability that the permutation
A) begins with a consonant
B) ends with a vowel
C) has the consonants and vowels alternating
please give a explanatiion
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website! if a permutation of the word "white" is selected at random,
find the probability that the permutation
The denominator of each of the probabilities will be 5P5 = 5! = 120
A) begins with a consonant
Pick the 1st letter any of 3 ways (w, h or t)
Arrange the remaining 4 letters in 4P4 = 4! = 24 ways
That's 3∙24 = 72 ways
Probability = 72/120 = 3/5
B) ends with a vowel
Pick the 5th letter either of 2 ways (i or e)
Arrange the remaining 4 letters in 4P4 = 4! = 24 ways
That's 2∙24 = 48 ways
Probability = 48/120 = 2/5
C) has the consonants and vowels alternating
CVCVC
The 3 consonants can be rearranged in the 1st, 3rd, and 5th
positions in 3P3 = 3! = 6 ways.
The 2 vowels can be rearranged in the 2nd and 4th
positions in 2P2 = 2! = 2 ways.
That's 6∙2 = 12 ways
Probability = 12/120 = 1/10
Edwin
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
In probability, it is useful to understand different methods for finding answers to questions like these. The answers from the other tutor are correct; and you should understand them and be able to use his methods.
His method is basically to find the number of favorable outcomes for each of the three cases and divide that by the total number of possible outcomes.
I would prefer a different general method for all three parts of the problem: finding the probabilities one letter at a time.
For the first two cases, that makes the problem very easy.
A) begins with a consonant: 3 of the 5 letters are consonants; there are no restrictions on the remaining letters. The probability is 3/5.
B) ends with a vowel: in the exact similar way, 2 of the 5 letters are vowels, and there are no other restriction; the probability is 2/5.
C) has the consonants and vowels alternating. For this case, we have to choose consonants for the 1st, 3rd, and 5th letters, and vowels for the 2nd and 4th.
P(1st a consonant) = 3/5
P(2nd a vowel) = 2/4
P(3rd a consonant) = 2/3
P(4th a vowel) = 1/2
P(5th a consonant) = 1/1
P(alternating consonants and vowels) = (3/5)(2/4)(2/3)(1/2)(1/1) = 12/120 = 1/10
The general methods used by me and the other tutor are both very basic. Again I state that you should know how to use both methods.
If you are just learning about probability, I would recommend trying both methods on each problem. Getting the same answer by two different methods gives you confidence that you are using the methods and doing the calculations correctly.
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