It is currently 19 Nov 2017, 19:32

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

SVP
Joined: 21 Jul 2006
Posts: 1512

Kudos [?]: 1051 [1], given: 1

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

### Show Tags

08 Dec 2007, 09:24
1
KUDOS
21
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

52% (01:32) correct 48% (01:28) wrong based on 611 sessions

### HideShow timer Statistics

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%
[Reveal] Spoiler: OA

Kudos [?]: 1051 [1], given: 1

Director
Joined: 30 Nov 2006
Posts: 591

Kudos [?]: 318 [0], given: 0

Location: Kuwait

### Show Tags

08 Dec 2007, 10:39
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

Kudos [?]: 318 [0], given: 0

Director
Joined: 12 Jul 2007
Posts: 857

Kudos [?]: 334 [2], given: 0

### Show Tags

08 Dec 2007, 10:44
2
KUDOS
1
This post was
BOOKMARKED
Mishari wrote:
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

.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%

Kudos [?]: 334 [2], given: 0

Senior Manager
Joined: 13 Jun 2007
Posts: 409

Kudos [?]: 83 [1], given: 0

Schools: Wharton, Booth, Stern

### Show Tags

20 Dec 2007, 02:50
1
KUDOS
1
This post was
BOOKMARKED
eschn3am wrote:
Mishari wrote:
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

.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%

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.

Kudos [?]: 83 [1], given: 0

CEO
Joined: 29 Mar 2007
Posts: 2553

Kudos [?]: 528 [0], given: 0

### Show Tags

20 Dec 2007, 22:25
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%

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

Kudos [?]: 528 [0], given: 0

Director
Joined: 03 Sep 2006
Posts: 864

Kudos [?]: 1106 [0], given: 33

### Show Tags

21 Dec 2007, 05:38
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%

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.

Kudos [?]: 1106 [0], given: 33

SVP
Joined: 07 Nov 2007
Posts: 1790

Kudos [?]: 1088 [1], given: 5

Location: New York

### Show Tags

25 Aug 2008, 12:45
1
KUDOS
1
This post was
BOOKMARKED
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%

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

Smiling wins more friends than frowning

Last edited by x2suresh on 27 Aug 2008, 23:45, edited 1 time in total.

Kudos [?]: 1088 [1], given: 5

Manager
Joined: 27 Oct 2008
Posts: 185

Kudos [?]: 166 [0], given: 3

### Show Tags

27 Sep 2009, 20:57
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

Ans is b

Kudos [?]: 166 [0], given: 3

Senior Manager
Joined: 22 Dec 2009
Posts: 356

Kudos [?]: 418 [0], given: 47

### Show Tags

16 Feb 2010, 08:10
1
This post was
BOOKMARKED
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%

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

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

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Kudos [?]: 418 [0], given: 47

Intern
Joined: 21 Feb 2010
Posts: 33

Kudos [?]: 7 [0], given: 9

Location: Ukraine

### Show Tags

25 Apr 2010, 03:17
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!

Kudos [?]: 7 [0], given: 9

Math Expert
Joined: 02 Sep 2009
Posts: 42259

Kudos [?]: 132730 [2], given: 12335

### Show Tags

25 Apr 2010, 04:08
2
KUDOS
Expert's post
3
This post was
BOOKMARKED
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 $$P=\frac{5!}{4!}*0.9^4*0.1$$.

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).
_________________

Kudos [?]: 132730 [2], given: 12335

Manager
Joined: 29 Dec 2009
Posts: 112

Kudos [?]: 26 [0], given: 10

Location: india

### Show Tags

01 May 2010, 00:39
it is like heads and tails question .... there if coin is biased then formula is

p of head^no. of heads * P. of tail ^ no. of tails * total toses Combination no. of head

so applying this v get ---- .9^4 * .1^1 * 5C4 = 32.8 %

Kudos [?]: 26 [0], given: 10

Manager
Joined: 07 Feb 2011
Posts: 105

Kudos [?]: 64 [0], given: 45

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

### Show Tags

22 Jan 2013, 03:23
Could someone please explain why they multiplied by 5 here? I don't get how order matters in this problem
_________________

Kudos [?]: 64 [0], given: 45

Math Expert
Joined: 02 Sep 2009
Posts: 42259

Kudos [?]: 132730 [0], given: 12335

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

### Show Tags

22 Jan 2013, 06:38
manimgoindowndown wrote:
Could someone please explain why they multiplied by 5 here? I don't get how order matters in this problem

Please go through this post: there-is-a-90-chance-that-a-registered-voter-in-burghtown-56812.html#p717422

Also, check Probability and Combinatorics chapters of our Math Book: gmat-math-book-87417.html
Probability: math-probability-87244.html;
Combinatorics: math-combinatorics-87345.html

Hope it helps.
_________________

Kudos [?]: 132730 [0], given: 12335

Manager
Joined: 07 Feb 2011
Posts: 105

Kudos [?]: 64 [0], given: 45

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

### Show Tags

22 Jan 2013, 16:01
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)
_________________

Kudos [?]: 64 [0], given: 45

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7736

Kudos [?]: 17806 [1], given: 235

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

### Show Tags

22 Jan 2013, 20:55
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17806 [1], given: 235 Manager Joined: 07 Feb 2011 Posts: 105 Kudos [?]: 64 [0], given: 45 Re: There is a 90% chance that a registered voter in Burghtown [#permalink] ### Show Tags 23 Jan 2013, 20:38 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? _________________ We appreciate your kudos' Kudos [?]: 64 [0], given: 45 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7736 Kudos [?]: 17806 [1], given: 235 Location: Pune, India Re: There is a 90% chance that a registered voter in Burghtown [#permalink] ### Show Tags 23 Jan 2013, 21:05 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/) I suggest you to go through the posts I have written on PnC on my blog to help you differentiate between Permutations and Combinations: http://www.veritasprep.com/blog/categor ... om/page/2/ 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17806 [1], given: 235

Manager
Joined: 07 Feb 2011
Posts: 105

Kudos [?]: 64 [0], given: 45

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

### Show Tags

23 Jan 2013, 23:05
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
_________________

Kudos [?]: 64 [0], given: 45

Non-Human User
Joined: 09 Sep 2013
Posts: 15659

Kudos [?]: 282 [0], given: 0

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

### Show Tags

29 Jul 2015, 06:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 282 [0], given: 0

Re: There is a 90% chance that a registered voter in Burghtown   [#permalink] 29 Jul 2015, 06:18

Go to page    1   2    Next  [ 25 posts ]

Display posts from previous: Sort by