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# There is a word AUSTRALIA. Four letters are taken at random.

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There is a word AUSTRALIA. Four letters are taken at random. [#permalink]

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10 Feb 2003, 03:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

There is a word AUSTRALIA. Four letters are taken at random. What is the probability to have two vowels and two consonants?
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10 Feb 2003, 04:08
With replacement or without?
_________________

Proud to be Azeri

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05 Jun 2003, 16:35
Can anybody post a solution to this one?
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05 Jun 2003, 21:30
vowels are A, U, A, I, and A.
consonants are S, T, R, and L.

total—9 letters

NO REPLACEMENT CASE:
Four are taken without replacement—9C4
2 vowels and 2 consonants—5C2*4C2 (remember that we do not need to find DIFFERENT combinations, so three As are OK)
Thus—it is 10/21

REPLACEMENT CASE:
P(V+V+C+C)=4C2*5/9*5/9*4/9*4/9=6*400/6561=6*0.06=0.36

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05 Jun 2003, 21:43
Thanks.

I needed it for the Database of questions.
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06 Jun 2003, 06:44
I have this solution

(5C2*2C4) / 4C9
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06 Jun 2003, 22:57
i got 2/15 (without repl). anyone else?
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06 Jun 2003, 23:07
Stolyar, help
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08 Jun 2003, 23:38
still need to brush up on prob...but my take:

(1/5)*(1/4) + (1/4)*(1/3) = 2/15
vowels consonants
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09 Mar 2005, 09:15
What is the OA for this one with replacement?
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09 Mar 2005, 09:37
Anonymous wrote:
There is a word AUSTRALIA. Four letters are taken at random. What is the probability to have two vowels and two consonants?

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of without replacement.
Without replacement:
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

But we can also consider the case of with replacement:
Total outcomes: 9^4
Outcomes with two vowels and two consonants: P(4,2)*5^2*4^2
You can calculate the probability from here.

Note that there is a special thing with letter problems, ie. the repeating letters. In the above question the repeating letters don't matter. However they would matter is the question is asked differently.

Example:
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?

Solution:
Vowels: A, U, A, I, A. Distinguish vowles: A, U, I.
Consonants: S, T, R, L

Outcomes with two vowels and two consonants:
1) The two vowels are different: C(3,2)*C(4,2)
2) The two vowels are the same: C(1,1)*C(4,2)
Total outcomes = 18+6=24
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09 Mar 2005, 12:41
Now thats a quality post...this should be a probability sticky
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09 Mar 2005, 14:47
Hmmm ok.
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09 Mar 2005, 14:58
HongHu wrote:
Anonymous wrote:
This is a case of without replacement.
Without replacement:
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21

But we can also consider the case of with replacement:
Total outcomes: 9^4
Outcomes with two vowels and two consonants: P(4,2)*5^2*4^2
You can calculate the probability from here.

Hong can u plz tell me what replacement means here, do you mean repetition????
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09 Mar 2005, 15:00
You take a letter, and then put it back before you take another.
09 Mar 2005, 15:00
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