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S1 probability distribution help

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1. The letters of the word DISTRIBUTION are written on separate cards. The cards are then shuffled and the top three are turned over.Let the random variable V be the number of vowels that are turned over.

(a) Show that P(V = 1) = 21/44 .

(b) Find the probability distribution of V.

I can prove part (a) but when I try to do part (b), the sum of my total probabilities do not equal 1.

So probability of a vowel= 5/12 which then becomes 4/11 as you pick another vowel etc
Probability of not a vowel= 7/12 which then becomes 6/11 etc

I worked out P(V = 0) = 21/44 from 3(7/12 x 6/11 x 5/10) ,
P(V = 1) = 21/44
P(V = 2) = 7/22
P(V = 3) = 3/22

But obviously these don't add up to 1 so it is incorrect :\ Can someone help and/or let me know if any of the probabilities are incorrect or maybe even the question is just dodgy? Idk..
2. (Original post by jessyjellytot14)
The letters of the word DISTRIBUTION are written on separate cards. The cards are then shuffled and the top three are turned over.Let the random variable V be the number of vowels that are turned over.

(a) Show that P(V = 1) = 21/44 .

(b) Find the probability distribution of V.

I can prove part (a) but when I try to do part (b), the sum of my total probabilities do not equal 1.

So probability of a vowel= 5/12 which then becomes 4/11 as you pick another vowel etc
Probability of not a vowel= 7/12 which then becomes 6/11 etc

I worked out P(V = 0) = 21/44 from 3(7/12 x 6/11 x 5/10) ,
Nearly right. But there is no requirement to multiply by 3.
If w represents a vowel and C a consonant, then you can only have CCC.

When V=1, you could have WCC or CWC, or CCW. Which is why you multiply by 3, having worked out the probability of 1 of the three combinations.

Similarly for V=2

With V=3 there is only one possibility WWW, so again no multiplying by 3.

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Updated: June 12, 2016
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