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# There is a 90% chance that a registered voter in Burghtown

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Manager
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There is a 90% chance that a registered voter in Burghtown [#permalink]  06 Nov 2004, 21:56
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There is a 90% chance that a registered voter in Burghtown voted in the last election. If five registered voters are chosen at random, what is the approximate likelihood that exactly four of them voted in the last election?

A) 26.2%

B) 32.8%

C) 43.7%

D) 59.0%

E) 65.6%

Thanks!
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B) 32.8
prob of person voting: 9/10
prob of person not voting: 1/10
(9/10)^4 * 1/10 * 5C4 = 32.8%
You have to multiply by 5C4 because that is the possible number of ways of arranging the 4 voting persons out of 5 picked
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Paul

Manager
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Paul wrote:
B) 32.8
prob of person voting: 9/10
prob of person not voting: 1/10
(9/10)^4 * 1/10 * 5C4 = 32.8%
You have to multiply by 5C4 because that is the possible number of ways of arranging the 4 voting persons out of 5 picked

Thanks Paul..That is where I got lost..I still don't get it why we need to multiply by 5C4. The questions asks the voting probability of 4 voters out of 5.. So why the order of the voters is important?

Last edited by afife76 on 06 Nov 2004, 22:21, edited 1 time in total.
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Alright, we know that there are 4(Y) who voted and 1(N) who did not.
We can have:
YYYYN
YYYNY
YYNYY
YNYYY
NYYYY
Hence, 5C4 = 5
You also have to take into account their respective probability of voting and not voting being 9/10 * 9/10 * 9/10 * 9/10 * 1/10(for the non-voter)
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Paul

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hey paul
wat is the answer if the question was ..... If five registered voters are chosen at random, what is the approximate likelihood that exactly three of them voted in the last election

is it 9/10^3 * 1/10^2 *5c3

jst clearing some doubts !
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Jim

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jimishg wrote:
hey paul
wat is the answer if the question was ..... If five registered voters are chosen at random, what is the approximate likelihood that exactly three of them voted in the last election

is it 9/10^3 * 1/10^2 *5c3

jst clearing some doubts !

_________________

Best Regards,

Paul

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