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

Question

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%

Please provide your steps when choosing your answer

A. 26.2%
B. 32.8%
C. 43.7%
D. 59.0%
E. 65.6%

Bunuel · Accepted Answer

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).

KarishmaB · Answer

The point is that the 5 voters are unique. Say they are A, B, C, D and E.
Now the case where A, B, C and D voted in the last election is different from the case where B, C, D and E voted in the last election. These are two different ways in which we can have 4 of the 5 who voted in the last election. There are 5 such different cases.

When you .9 * .9 * .9 * .9 * .1, you are finding the probability that A, B, C and D voted in the last election while E did not.
So you have to count 4 more cases e.g. .1 * .9 * .9 * .9 * .9 (Probability that A did not vote in the last election while B, C, D , E voted in the last election) etc.
The calculation is identical to the first case so you just need to count the first case 5 times.

Mishari · Answer

Actually this is not really a probability question.

The probability that four of five voted is :
P(1st one voted) X ... X P(4th one voted) X (5th one NOT voted)
= 0.9 x 0.9 x 0.9 x 0.9 x 0.1
= 0.81 x 0.81 x 0.1 =  0.6561 


 ANSWER:   E

eschn3am · Answer

.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

Audio · Answer

Another straightforward way is to applicate the probability formula:

C(4,5) x (9/10)^4 x (1/10)^1 = 5 x 0.81 x 0.81 x 0.1. 

No room for mistakes in this case.

x2suresh · Answer

=5C4* (0.9)^4 * (O.1)^1

srivas · Answer

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%

Please provide your steps when choosing your answer

a) 26.2%
b) 32.8%
c) 43.7%
d) 59.0%
e) 65.6%

Soln: 
Probability that voter votes = .9
Probability that voter does not vote = .1

Let Y be that voter votes and N be that voter does not vote
Therefore we have YYYYN

Probability that exactly 4 out of 5 vote 
= (5!/4!) * (.9)^4 * (.1)^1
= .328

Ans is b

jeeteshsingh · Answer

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

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

fruit · Answer

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

manimgoindown · Answer

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)

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)

manimgoindown · Answer

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?

KarishmaB · Answer

The committee doesn't have unique positions - team lead, member etc. The question is that of the total 3 member committees you can form out of the 6 members, how many that have Mike have Anthony as well. It is a combinations question. No of committees that have Mike = 5C2 = 10 (you include Mike and then choose 2 members out of the other 5) No of committees that have Mike have Anthony too = 4C1 = 4 (you include Mike and Anthony and choose 1 member of the remaining 4) Probability = 4/10 or when you use probability, you write the probability of the next move. You do not make two moves at the same time. So you can pick Anthony when you pick the second member or you can pick him when you pick the third member. The probability of picking Anthony in each case is 1/5 and you add them up to get 2/5. (Why will the probability stay the same 1/5 in both cases? Check out this video on Probability: https://youtu.be/0BCqnD2r-kY

ScottTargetTestPrep · Answer

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

bumpbot · Answer

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

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