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Four letters are taken at random from the word AUSTRALIA.

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Director
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Four letters are taken at random from the word AUSTRALIA. [#permalink]

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15 Jul 2006, 01:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

1. Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

2. From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen?

3. There is a word AUSTRALIA. Four letters are taken at random. How many different words can be formed that have two vowels and two consonants?
Director
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15 Jul 2006, 05:54
#1. There are 5 vowels and 4 consonants. Assuming no replacement, probability should be 5C2*4C2/9C4.

#2. Since there is replacement, probability of a vowel is 5/9 and probability of a consonant is 4/9. Probability should be [(5/9)^2]*[(4/9)^2]

#3. Working on this one...
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Uh uh. I know what you're thinking. "Is the answer A, B, C, D or E?" Well to tell you the truth in all this excitement I kinda lost track myself. But you've gotta ask yourself one question: "Do I feel lucky?" Well, do ya, punk?

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22 Jul 2006, 03:02
#2: You need to multiply that figure by 6 to get the probability.
VVCC
VCVC
VCCV
CVVC
CCVV
CVCV
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22 Jul 2006, 12:52
I will go like this
Two vovels and two consenants can be selected as 5C2*4C2 ways.
Assumptions : Word will not be meaning full(if this assumption is wrong then do it will be better to guess on this in actual exam rather than trying it).
Then the 4 letter (two vovel and 2 consants) can be arranged in 4^4 ways to form words
So the ans is 5C2*4C2*(4^4) ways.

Pravin
#3   [#permalink] 22 Jul 2006, 12:52
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