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

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There is a 90% chance that a registered voter in Burghtwon [#permalink] New post 30 Oct 2005, 22:14
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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Re: PR Test 2 # 20 Probablility [#permalink] New post 30 Oct 2005, 22:17
=5c4 (0.9)^4(0.1)
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 [#permalink] New post 30 Oct 2005, 22:26
same as Himalaya use binomial

ACB P^(B)*Q(A-B)
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 [#permalink] New post 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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 [#permalink] New post 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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 [#permalink] New post 31 Oct 2005, 07:39
Answer is B.
--

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%
  [#permalink] 31 Oct 2005, 07:39
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