Posted by **John** on Wednesday, March 3, 2010 at 5:32pm.

I'm looking at my notes and I can't seem to figure how my teacher did this problem. She used her name which has 9 letters, 3 vowels, and 6 consonants. If we were to reach in the bag and grab 3 letters out. By using the probability distribution what's the probability of getting 0, 1, 2, and 3 vowels. I know how to do the probability of 0 vowel. For 1 vowel she did (3/9*6/8*5/7)3 how did she get 6/8 and 5/7. For 2 vowels she did (3/9*2/8*6/7)3 same here where did the 2/8 and 6/7 come from. Then for 3 vowels she had 1 minus everything.

- Statistics -
**Reiny**, Wednesday, March 3, 2010 at 6:09pm
I will do the 2 vowels for you

since she grabbed 3 letters it could have been

VVC, VCV, or CVV

look at the first, the VVC

there are 3 vowels out of 9 prob (first a vowel) = 3/9

so now there are 8 letters left, 2 of them vowels

prob (second a vowel) = 2/8

so now the third letter must be a consonant, and there are 7 letters left, 6 of them consonants.

Prob(third is a consonant) = 6/7

so

Prob(first a vowel, 2nd a vowel, third a cons) = 3/9 x 2/8 x 6/7

now can you see how she would get the VCV of

3/9 x 6/8 x 2/7 setup ?

and for the CVV it would be 6/9 x 3/8 x 2/7

did you notice that the numerators are simply

2x3x6 and the denominators are 9x8x7 just in different order?

Does it matter what order you do multiplication?

so the prob would be

(3/9*2/8*6/7)(3/9*2/8*6/7)(3/9*2/8*6/7)3

=(3/9*2/8*6/7)^3

Now see if you reason out the others in the same way.

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