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

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Director
Joined: 15 Aug 2005
Posts: 804
Location: Singapore
Followers: 2

Kudos [?]: 18 [0], given: 0

There is a 90% chance that a registered voter in Burghtown [#permalink]  18 Sep 2005, 00:00
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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 excatly four of them voted in the last election?

A. 26.2%
B. 32.8%
C. 43.7%
D. 59.0%
E. 65.6%

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Cheers, Rahul.

Manager
Joined: 14 Jul 2005
Posts: 104
Location: Sofia, Bulgaria
Followers: 1

Kudos [?]: 7 [0], given: 0

Once you identify that this is binomial distribution case, you should simply use the formula.

P = (0.9^4) * (0.1^1) * C(5,4) = 32.8%

Narrative: I assume you just multiplied 0.9 four times to reach the answer. That's not enough, because this way you can't be sure about the voting status of the 5th person. So you plug in his 10% probability, and you get (0.9^4) * (0.1^1) ~ 0.066. But this is incomplete, because we haven't considered if the 10% person shows up 1st, 2nd, etc. In order to stir things up, we multiply by C(5,4) = C(5,1) to account for all possible positions of the non-voter among the voters.
Intern
Joined: 30 Aug 2005
Posts: 9
Followers: 0

Kudos [?]: 3 [0], given: 0

Yes, agreed with vasild. Using Binomial distribution is the quickest way to solve this problem.

Intern
Joined: 19 Aug 2005
Posts: 41
Followers: 0

Kudos [?]: 1 [0], given: 0

got B too using binomial distribition also
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