Find all School-related info fast with the new School-Specific MBA Forum

It is currently 25 Oct 2014, 17:51

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
SVP
SVP
avatar
Joined: 21 Jul 2006
Posts: 1551
Followers: 8

Kudos [?]: 236 [0], given: 1

There is a 90% chance that a registered voter in Burghtown [#permalink] New post 08 Dec 2007, 08:24
4
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

46% (02:31) correct 54% (01:24) wrong based on 213 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?

A. 26.2%
B. 32.8%
C. 43.7%
D. 59.0%
E. 65.6%
[Reveal] Spoiler: OA
Director
Director
User avatar
Joined: 30 Nov 2006
Posts: 592
Location: Kuwait
Followers: 11

Kudos [?]: 154 [0], given: 0

 [#permalink] New post 08 Dec 2007, 09:39
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


ANSWER: E
Director
Director
User avatar
Joined: 12 Jul 2007
Posts: 867
Followers: 12

Kudos [?]: 202 [0], given: 0

 [#permalink] New post 08 Dec 2007, 09:44
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


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
Senior Manager
Senior Manager
User avatar
Joined: 13 Jun 2007
Posts: 410
Schools: Wharton, Booth, Stern
Followers: 11

Kudos [?]: 79 [0], given: 0

 [#permalink] New post 20 Dec 2007, 01:50
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


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.

No room for mistakes in this case.
CEO
CEO
User avatar
Joined: 29 Mar 2007
Posts: 2593
Followers: 16

Kudos [?]: 198 [0], given: 0

Re: PS: Probability [#permalink] New post 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...

5*.06561=~.328

B
Director
Director
User avatar
Joined: 03 Sep 2006
Posts: 889
Followers: 6

Kudos [?]: 171 [0], given: 33

Re: PS: Probability [#permalink] New post 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.
SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1829
Location: New York
Followers: 27

Kudos [?]: 472 [0], given: 5

Re: PS: Probability [#permalink] New post 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.
Manager
Manager
avatar
Joined: 27 Oct 2008
Posts: 186
Followers: 1

Kudos [?]: 80 [0], given: 3

Re: PS: Probability [#permalink] New post 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

Ans is b
Senior Manager
Senior Manager
User avatar
Joined: 22 Dec 2009
Posts: 364
Followers: 10

Kudos [?]: 217 [0], given: 47

GMAT ToolKit User
Re: PS: Probability [#permalink] New post 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
_________________

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

|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|


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

Intern
Intern
User avatar
Joined: 21 Feb 2010
Posts: 33
Location: Ukraine
Followers: 1

Kudos [?]: 3 [0], given: 9

Re: PS: Probability [#permalink] New post 25 Apr 2010, 02:17
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!
Expert Post
2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23422
Followers: 3618

Kudos [?]: 28976 [2] , given: 2874

Re: PS: Probability [#permalink] New post 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!.

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).
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 29 Dec 2009
Posts: 122
Location: india
Followers: 1

Kudos [?]: 14 [0], given: 10

Re: PS: Probability [#permalink] New post 30 Apr 2010, 23:39
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 %
Manager
Manager
User avatar
Joined: 07 Feb 2011
Posts: 89
Followers: 0

Kudos [?]: 20 [0], given: 43

Re: There is a 90% chance that a registered voter in Burghtown [#permalink] New post 22 Jan 2013, 02:23
Could someone please explain why they multiplied by 5 here? I don't get how order matters in this problem
_________________

We appreciate your kudos'

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23422
Followers: 3618

Kudos [?]: 28976 [0], given: 2874

Re: There is a 90% chance that a registered voter in Burghtown [#permalink] New post 22 Jan 2013, 05:38
Expert's post
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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 07 Feb 2011
Posts: 89
Followers: 0

Kudos [?]: 20 [0], given: 43

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

We appreciate your kudos'

Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4877
Location: Pune, India
Followers: 1157

Kudos [?]: 5383 [1] , given: 165

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

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Manager
User avatar
Joined: 07 Feb 2011
Posts: 89
Followers: 0

Kudos [?]: 20 [0], given: 43

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

We appreciate your kudos'

Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4877
Location: Pune, India
Followers: 1157

Kudos [?]: 5383 [1] , given: 165

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

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

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Manager
User avatar
Joined: 07 Feb 2011
Posts: 89
Followers: 0

Kudos [?]: 20 [0], given: 43

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

We appreciate your kudos'

Re: There is a 90% chance that a registered voter in Burghtown   [#permalink] 23 Jan 2013, 22:05
    Similar topics Author Replies Last post
Similar
Topics:
. In a certain city, 60 percent of the registered voters are banksy 3 18 Mar 2011, 13:35
There is a 90% chance that a registered voter in Burghtwon joemama142000 5 30 Oct 2005, 21:14
There is a 90% chance that a registered voter in Burghtown rahulraao 3 18 Sep 2005, 00:00
There is a 90% chance that a registered voter in Burghtown gayathri 2 30 Nov 2004, 09:34
There is a 90% chance that a registered voter in Burghtown afife76 5 06 Nov 2004, 21:56
Display posts from previous: Sort by

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

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.