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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, 21:01
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, 09:48
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, 04:53
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.
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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, 15:39
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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Re: There is a 90% chance that a registered voter in Burghtown  [#permalink]

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New post 13 Jan 2018, 12:22
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%



v v v v n

(9/10)^4 * (1/10)^1 * 5

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

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New post 15 Jun 2018, 00:47
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%


Probability of a registered voter, voted in last election = 90%

So probability when 5 registered voters are selected, exactly 4 of them voted in last election = VVVVN = (90/100)^4*(10/100)*5!/4! = 0.32805 = 32.8%

Answer B.


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

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New post 02 Aug 2019, 02:40
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%

Chance of voting is 90% i.e. 0.9
Chance of not voting becomes 10% i.e. 0.1

four out of five votes then, 0.9 x 0.9 x 0.9 x 0.9 x 0.1 and this can be arranged in itself in 5!/4! ways ie. 5
hence total will be 0.32805 or 32.8 %

B is the answer.

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

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New post 26 Aug 2019, 03:20
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%


total exactly 4 voted out of 5 ; 5c4 * ( .9)^4*(.1) = 32.8%
IMO B
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Re: There is a 90% chance that a registered voter in Burghtown  [#permalink]

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New post 26 Aug 2019, 03:40
(0.9)^4*0.1* 5C4=
6561/20000=~32.8%
Option B

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

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New post 16 Jun 2020, 02:06
Bunuel wrote:
fruit wrote:
jeeteshsingh wrote:
tarek99 wrote:


VVVVN.... V represent voted... and N represent Not Voted.

Therefore % = .9 * .9 * .9 * .9 * .1 x 5!/4! x100 = 32.8% = B


Is there any general formula?
I can't get for what do you do 5!/4!


If the probability of a certain event is \(p\), then the probability of it occurring \(k\) times in \(n\)-time sequence is: \(P = C^k_n*p^k*(1-p)^{n-k}\)

In our case:
n=5
k=4
p=0.9

So, \(P = C^k_n*p^k*(1-p)^{n-k}=C^4_5*0.9^4*0.1\)

OR: probability of scenario V-V-V-V-N is \(0.9^4*0.1\), but V-V-V-V-N can occur in different ways:
V-V-V-V-N - first four voted and fifth didn't;
N-V-V-V-V - first didn't vote and last four did;
V-N-V-V-V first voted, second didn't and last three did;
...

Certain # of combinations. How many combinations are there? Basically we are looking at # of permutations of five letters V-V-V-V-N, which is 5!/4!.

Hence \(P=\frac{5!}{4!}*0.9^4*0.1\).


Also you can check Probability chapter of Math Book for more (link in my signature).



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Hi -just out of interest on this - what would be your best strategy for actually calculating the final percentage from the numbers given. Like how would you best do the difficult calculation or estimate the answer without wasting lots of time? Many thanks :)
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Re: There is a 90% chance that a registered voter in Burghtown   [#permalink] 16 Jun 2020, 02:06

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