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# This topic was discussed before.. There is a word AUSTRALIA.

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Senior Manager
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This topic was discussed before.. There is a word AUSTRALIA. [#permalink]  05 Jul 2006, 13:32
This topic was discussed before..

There is a word AUSTRALIA. Four letters are taken at random. What is the probability to have two vowels and two consonants? Consider the case with repalcement..

http://www.gmatclub.com/phpbb/viewtopic ... robability
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Manager
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[#permalink]  05 Jul 2006, 18:32
I've got 30/63.

There are 5 vowels and 4 consonants in th word AUSTRALIA.

Total number of combinations of 4 letters: 9C4 = 126

2 vowels and 2 consonants can be selected in 2C5*2C4 = 60 ways

So, Prob = 60/126 = 30/63
SVP
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[#permalink]  05 Jul 2006, 22:41
another way (Edit : Sorry considered it with replacement)

Four letters ( two vowels and two consonants) can be chosen in 6 ways

VVCC
VCVC
CVCV
CCVV
VCCV
CVVC

Each has a probability of 5/9*4/8*4/7*3/6 = 5/63

Hence total P = 6*5/63 = 30/63 = 10/21
SVP
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[#permalink]  05 Jul 2006, 22:50
Without replacement

Same approach
Just that probabilty of individual cases become 5/9*5/9*4/9*4/9 = 5^2*4^2/9^4

hence total P = 6* 5^2 * 4^2 / 9^4
CEO
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[#permalink]  06 Jul 2006, 07:20
Prob of selcting a vowel = 5/9
Prob of selecting a consonant = 4/9

P (VVCC) = 5/9 * 5/9 * 4/9 * 4/9
Total combinations of VVCC (VVCC, CCVV, CVCV, VCVC, VCCV, CVVC) = = 4!/ (2! * 2!) = 6

So prob = 6 * 5/9 * 5/9 * 4/9 * 4/9
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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[#permalink]  06 Jul 2006, 18:34
With replacement
vowels = A,A,I,A --> 4 of them
consonants = u,s,t,r,l --> 5 of them

We want to end up with 2 vowels and 2 consonants. So we can write a word VVCC, and the number of ways we can have 2 vowels and 2 consonants = 4!/2!2! = 6 ways

i.e.
VVCC --> Probability = (1/4)^2(1/5)^2 = 1/400
VCVC --> Probability = (1/4)^2(1/5)^2 = 1/400
CCVV --> Probability = (1/4)^2(1/5)^2 = 1/400
CVCV --> Probability = (1/4)^2(1/5)^2 = 1/400
VCCV --> Probability = (1/4)^2(1/5)^2 = 1/400
CVVC --> Probability = (1/4)^2(1/5)^2 = 1/400

Total P = 1/400 * 6 = 3/200
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[#permalink]  06 Jul 2006, 18:49
wilfred - why are the individual probabilities 1/4 and 1/5 when we can replace the letters? Kindly explain.
_________________

Thanks,
Zooroopa

[#permalink] 06 Jul 2006, 18:49
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# This topic was discussed before.. There is a word AUSTRALIA.

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