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

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joemama142000
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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: There is a 90% chance that a registered voter in Burghtwon [#permalink] New post   30 Oct 2005, 22:17
=5c4 (0.9)^4(0.1)
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Re: There is a 90% chance that a registered voter in Burghtwon [#permalink] New post   30 Oct 2005, 22:26
same as Himalaya use binomial

ACB P^(B)*Q(A-B)
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Re: There is a 90% chance that a registered voter in Burghtwon [#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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Re: There is a 90% chance that a registered voter in Burghtwon [#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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Re: There is a 90% chance that a registered voter in Burghtwon [#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%



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Re: There is a 90% chance that a registered voter in Burghtwon [#permalink]
31 Oct 2005, 07:39
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