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

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Director
Joined: 17 Oct 2005
Posts: 924

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There is a 90% chance that a registered voter in Burghtwon [#permalink]

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30 Oct 2005, 22:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

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SVP
Joined: 05 Apr 2005
Posts: 1708

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Re: PR Test 2 # 20 Probablility [#permalink]

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30 Oct 2005, 22:17
=5c4 (0.9)^4(0.1)

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Manager
Joined: 20 Mar 2005
Posts: 201

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Location: Colombia, South America

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30 Oct 2005, 22:26
same as Himalaya use binomial

ACB P^(B)*Q(A-B)

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Director
Joined: 17 Oct 2005
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30 Oct 2005, 22:31
thanks guys, however i don't understand the equation that you guys used. Where did it come from and when do you use such an equation?

What is ACB?
and 5c4?

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Manager
Joined: 20 Mar 2005
Posts: 201

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Location: Colombia, South America

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30 Oct 2005, 22:41
joemama142000 wrote:
thanks guys, however i don't understand the equation that you guys used. Where did it come from and when do you use such an equation?

What is ACB?
and 5c4?

sorry about my notation it is certainly pretty lame

AcB means combinations, so A! / B! (A-B)!

p the probability of ocurrence and q would be (1-p)

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Intern
Joined: 30 Oct 2005
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31 Oct 2005, 07:39
--

1. Find combined probability of 4 out of 5 voters:
9/10 + 9/10 + 9/10 + 9/10 + 1/10 = 9^4/10^5

2. Find # of combinations when 4 out of 5 voters:
(5*4*3*2*1)/(4*3*2*1) = 5

3. Multiply # of Comb. by Probability:
5 * (9^4/10^5) = 32.8%

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31 Oct 2005, 07:39
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