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%

.81 * .81 * .1 = .06561, which isn't one of the choices.

I think that this formula represents that probability that the first 4 people will have voted and the 5th voter hasn't. Since we don't care what the order is we have to do this equation 5 times (shifting the .1 to each possible spot). Easier way, just multiply .06561 by 5 and you get...

32.8%

Answer B for me

We would have .9^4*.1 =.06561 however

we have 5 possiblities of arraganging the four .9's and the .1 so...

5*.06561=~.328

B

=5C4* (0.9)^4 * (O.1)^1
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VVVVN.... V represent voted... and N represent Not Voted.

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

Check this links for similar problems:
combinatorics-at-least-none-56728.html
probability-qs-attention-88945.html#p671944
probability-q-91460.html#p666937
permutation-86687.html#p650790

Also you can check Probability chapter of Math Book for more (link in my signature).
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Could someone please explain why they multiplied by 5 here? I don't get how order matters in this problem
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Hey I still don't get it. The reason I ask is because the order of voting doesn't seem to matter overtly to me. We have those who vote and those who didn't. Why are we even considering the total number of permutations or combinations and multiplying it with the chance of voting when all that was asked was was the probability of all four voting. Since all of them are voting how does order matter here?

V V V V 4 to me there aren't five ways to arrange those votes because a vote is just a vote. Vote 1 is no different from Vote 2?

Basically I would select E, NOT B (which you get by multipling E by 5)
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I understand our explanation for that problem. Actually the more I do probability problems like I have been doing last week I find the biggest and most fundamental conceptual problem I have is when to include order/disclude order.

Let's take the following example

Anthony and Michael sit on the six member board of directors for company X. If the board is to be split up into 2 three- person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
question stem limits this problem very nicely.

(A) 20%
(B) 30%
(C) 40%
OA: C

The answer is C. Let's look at a committee where Mike belongs (he's person # 1) on that committee. When we choose person #2, there is a 1/5 probability that it's going to be Anthony (then it doesn't matter who the third person is), and 4/5 probability that it's going to be someone else (then person #3 is going to be Anthony with probability 1/4). Total probability = 1/5+4/5*1/4 = 2/5



Now I wonder why wouldn't we take 2/5 and multiply is by 4C4 as the user who solved it above did not include the combination in his calculation. I know the answer would be the same, but isn't the 4C4 a way of showing that the four positions that don't include Mike could be rearranged in so many ways, since order doesn't matter?
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Also I think a better way to think about this is not so much that the five voters are unique, but that each INSTANCE is unique
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