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

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Re: There is a 90% chance that a registered voter in Burghtown [#permalink]

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New post 14 Mar 2016, 18:22
tarek99 wrote:
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%


we need exactly 4 yes and 1 no.
5C4 = 5.
now, each combination is:
0.9x0.9x0.9x0.9x0.1 = 6.561
since we have 4 such options, multiply this number by 5.
round first to 6.56 x 5 = 32.8%

B

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Re: There is a 90% chance that a registered voter in Burghtown [#permalink]

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New post 14 Mar 2016, 22:01
tarek99 wrote:
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%

four can be selected from five in 5 ways. Each of the selection has a probability .9^4 *.1^1
Multiplying this with 5 we get the answer as .32805, which is 32.8%

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Re: There is a 90% chance that a registered voter in Burghtown [#permalink]

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New post 18 Mar 2016, 10:48
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?

Following are the cases:

P1(Voted )........P2(voted).........p3(voted)........ p4(voted).............p5(Not voted) = 0.9*0.9*0.9*0.9*0.1
P1(Voted )........P2(Not voted).........p3(voted)........ p4(voted).............p5(voted) = 0.9*0.1*0.9*0.9*09
P1(Voted )........P2(voted).........p3(Not voted)........ p4(voted).............p5(voted) = 0.9*0.9*0.1*0.9*0.9
P1(Voted )........P2(voted).........p3(voted)........ p4(Not voted).............p5(voted) = 0.9*0.9*0.9*0.1*0.9
P1(Not Voted )........P2(voted).........p3(voted)........ p4(voted).............p5(Not voted) = 0.1*0.9*0.9*0.9*0.9
Adding 5*(0.9)^4*0.1 = 32.8%

Answer is B

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 Burghtown [#permalink]

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New post 21 Sep 2017, 05:53
Hello Bunuel ,
What should I be looking for, in such question, to know whether I need to apply the combination or not.
What subtle details in the question stem would differentiate it from the normal one? Could you please elaborate a little.
_________________

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
When nothing seem to help, I would go and look at a Stonecutter hammering away at his rock perhaps a hundred time without as much as a crack showing in it.
Yet at the hundred and first blow it would split in two.
And I knew it was not that blow that did it, But all that had gone Before
.

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Re: There is a 90% chance that a registered voter in Burghtown [#permalink]

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New post 25 Sep 2017, 16:39
tarek99 wrote:
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%


We need to determine the probability that 4 voters vote and 1 does not. If we let Y = a voter voted and N = a voter didn’t vote, then:

P(Y-Y-Y-Y-N) = 0.9 x 0.9 x 0.9 x 0.9 x 0.1 = 0.06561

We also have to consider that we can arrange these 5 voters in 5!/4! = 5 ways; thus, the overall probability is:

5 x 0.06561 = 0.32805 = 32.8%

Answer: B
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Scott Woodbury-Stewart
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Re: There is a 90% chance that a registered voter in Burghtown   [#permalink] 25 Sep 2017, 16:39

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