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There is a 90% chance that a registered voter in Burghtown [#permalink]
08 Dec 2007, 08:24

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

45% (02:40) correct
54% (01:23) wrong based on 152 sessions

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?

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

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

.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...

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

.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

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.

Re: PS: Probability [#permalink]
20 Dec 2007, 21:25

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%

Please provide your steps when choosing your answer

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

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

Re: PS: Probability [#permalink]
21 Dec 2007, 04:38

GMATBLACKBELT wrote:

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%

Please provide your steps when choosing your answer

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

we have 5 possibilities of rearranging the four .9's and the .1 so...

5*.06561=~.328

B

I like this approach more, this is more logical and more clear. BUT FOR SURE, I had chosen the answer as Mishari did and I did not multiply the result ( 0.06561 ) with 5 different ways of arranging. Arrangement should not have mattered if all the 5 people were selected simultaneously!!!! But as the question does not mention that all 5 people are chosen simultaneously, why do we assume that they are chosen on-by-one and then of course raising the possibility of 5 different arrangements.

Re: PS: Probability [#permalink]
25 Aug 2008, 11:45

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%

Please provide your steps when choosing your answer

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

Your attitude determines your altitude Smiling wins more friends than frowning

Last edited by x2suresh on 27 Aug 2008, 22:45, edited 1 time in total.

Re: PS: Probability [#permalink]
27 Sep 2009, 19:57

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%

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

Re: PS: Probability [#permalink]
16 Feb 2010, 07:10

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%

Please provide your steps when choosing your answer

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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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: PS: Probability [#permalink]
25 Apr 2010, 03:08

2

This post received KUDOS

Expert's post

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!.

Re: There is a 90% chance that a registered voter in Burghtown [#permalink]
22 Jan 2013, 15:01

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)
_________________

Re: There is a 90% chance that a registered voter in Burghtown [#permalink]
22 Jan 2013, 19:55

1

This post received KUDOS

Expert's post

manimgoindowndown wrote:

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)

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.
_________________

Re: There is a 90% chance that a registered voter in Burghtown [#permalink]
23 Jan 2013, 19:38

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?
_________________

Re: There is a 90% chance that a registered voter in Burghtown [#permalink]
23 Jan 2013, 20:05

1

This post received KUDOS

Expert's post

manimgoindowndown wrote:

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?

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 here: http://www.veritasprep.com/blog/2012/10 ... ure-again/)

Go down the page and start with the post titled - The Dreaded Combinatorics Then go from down up and cover the PnC and Probability posts.
_________________

Re: There is a 90% chance that a registered voter in Burghtown [#permalink]
23 Jan 2013, 22:05

VeritasPrepKarishma wrote:

manimgoindowndown wrote:

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)

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.

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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Re: There is a 90% chance that a registered voter in Burghtown
[#permalink]
23 Jan 2013, 22:05