GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Dec 2018, 18:22

### 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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

December 14, 2018

December 14, 2018

10:00 PM PST

11:00 PM PST

Carolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.
• ### Free GMAT Strategy Webinar

December 15, 2018

December 15, 2018

07:00 AM PST

09:00 AM PST

Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.

# Out of 100 lottery tickets, each having 50% chance of

Author Message
Senior Manager
Joined: 13 Mar 2007
Posts: 276
Location: Russia, Moscow
Out of 100 lottery tickets, each having 50% chance of  [#permalink]

### Show Tags

06 Aug 2007, 09:15
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:04) correct 0% (00:00) wrong based on 13 sessions

### HideShow timer Statistics

Out of 100 lottery tickets, each having 50% chance of winning, 2 tickets are bought at random. What is the probability that at least one ticket will win?
(A) 1/4
(B) 1/2
(C) 49/198
(D) 149/198
(E) 3/4

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Intern
Joined: 06 Aug 2007
Posts: 14

### Show Tags

06 Aug 2007, 09:49
i think it goes like this:

total number of outcomes = 4 (lose-lose, win-win, win-lose, lose-win)
total number of outcomes where at least 1 ticket wins = 3

This one is a little tricky though, since they remaining 98 tickets you did not pick have no impact on the answer to the question. At least, i dont think they do....
Intern
Joined: 06 Aug 2007
Posts: 14

### Show Tags

06 Aug 2007, 10:14
hmm, I am not sure how you get D as the answer. Maybe someone else will help us on this one.

I've noticed a few other posts where the OA was wrong, perhaps this is one of them.
Intern
Joined: 06 Aug 2007
Posts: 37

### Show Tags

06 Aug 2007, 14:19
The way we can pick two tickets from the 100 available is 100C2 way = 50*99

Atleast one of them wins = 1 - both failing

Both failing = ???

How do we calculate the probablity of both of them failing from the selection?
Any help here is much appreciated!
VP
Joined: 10 Jun 2007
Posts: 1357

### Show Tags

06 Aug 2007, 15:18
Kalyan wrote:
The way we can pick two tickets from the 100 available is 100C2 way = 50*99

Atleast one of them wins = 1 - both failing

Both failing = ???

How do we calculate the probablity of both of them failing from the selection?
Any help here is much appreciated!

I got D too, don't know what's wrong...
both failing = (first fail) * (second fail) = 1/2 * 1/2 = 1/4
So Probability that at least one will win = 1-1/4 = 3/4.
Maybe someone else can explain how to get the OA.
Obviously, I agree that the total doesn't affect the outcome.
Senior Manager
Joined: 14 Jun 2007
Posts: 392

### Show Tags

06 Aug 2007, 18:03
bkk145 wrote:
Kalyan wrote:
The way we can pick two tickets from the 100 available is 100C2 way = 50*99

Atleast one of them wins = 1 - both failing

Both failing = ???

How do we calculate the probablity of both of them failing from the selection?
Any help here is much appreciated!

I got D too, don't know what's wrong...
both failing = (first fail) * (second fail) = 1/2 * 1/2 = 1/4
So Probability that at least one will win = 1-1/4 = 3/4.
Maybe someone else can explain how to get the OA.
Obviously, I agree that the total doesn't affect the outcome.

you got E you mean right? so did I. this is my kinda lotto!

every lotto ticket has a 1/2 chance of winning. its stated in the stem.

PR at least one = 1 - Pr none = 3/4
Intern
Joined: 11 Sep 2006
Posts: 45

### Show Tags

07 Aug 2007, 12:03
1
out of 100 each ticket has 1/2 probability of winning. So we have 50 winning and 50 non-winning tickets.

Now selecting 2 out of non-winning.
50/100 * 49/99

= 49/198 probabillity of non -winning tickets

Atleast one winning means 1-49/198 = 149/198
D.

Hope this helps.
Senior Manager
Joined: 14 Jun 2007
Posts: 392

### Show Tags

07 Aug 2007, 15:53
ajisha wrote:
out of 100 each ticket has 1/2 probability of winning. So we have 50 winning and 50 non-winning tickets.

Now selecting 2 out of non-winning.
50/100 * 49/99

= 49/198 probabillity of non -winning tickets

Atleast one winning means 1-49/198 = 149/198
D.

Hope this helps.

i have never taken a class in probability theory, but I have been reading up on it and is "each ticket has a 1/2 chance of winning" necessarily the same thing as "there are 50 winning tickets"

i'm not so keen on that. on the surface you can say "sure it is" but think about the binomial distribution of flipping a coin 100 times. you've got outliers where you can have 100 heads and 100 tails.

100 tickets each with 1/2 chance of winning means that it is possible for all 100 tickets to win. likewise it is possible for none of the tickets to be winners.

the question stem should say there are 100 lotto tickets, of which 50 are winning tickets. the difference may seem subtle, but implications on the correct calculation are not.

IMHO, this question is junk. If you approach it from the angle presented, it is completely illogical. If there are a 100 tickets; 50 of which are winners, each ticket DOES NOT have a 1/2 chance of winning. whether a ticket is a winner or loser has already been determined. Therefore the ticket IS A WINNER (P=1) or IS NOT A WINNER (P=0). Chance does not come into play whether it is a winner or a loser, it either is or isn't. If this is the case, then the answer is D. but the way the question is written is a mess.

It is hard to explain, but sit and think about it what i am saying for a while.
Intern
Joined: 06 Aug 2007
Posts: 14

### Show Tags

08 Aug 2007, 06:49
anonymousegmat, I agree with you strongly. You put it very very well.

Are you sure you've never taken a prob class before ?
Intern
Joined: 11 Sep 2006
Posts: 45

### Show Tags

08 Aug 2007, 08:44
Ok. So let us take it this way.

out of 100 tickets there are 50 winning and 50 loosing. if i select one out of it what is the probability of winning?

50/100 = 1/2. So it says its has 50% chance to win. It also implies that you might not win too. and that is why it is probability.

I just tried it by applying reverse logic, dont understand why it is not right way to solve.
Director
Joined: 12 Jul 2007
Posts: 843

### Show Tags

08 Aug 2007, 08:46
Quote:
out of 100 tickets there are 50 winning and 50 loosing

I only just skimmed this thread, but I believe the argument is being that you cannot assume this. They just say there are 100 tickets and each has a 50% chance of winning. That doesn't mean there are 50 winning and 50 losing tickets.
Senior Manager
Joined: 04 Jun 2007
Posts: 339

### Show Tags

08 Aug 2007, 08:58
ajisha wrote:
Ok. So let us take it this way.

out of 100 tickets there are 50 winning and 50 loosing. if i select one out of it what is the probability of winning?

50/100 = 1/2. So it says its has 50% chance to win. It also implies that you might not win too. and that is why it is probability.

I just tried it by applying reverse logic, dont understand why it is not right way to solve.

If there are 100 tickets with each having 50% chance of winning, it cannot be assumed there are 50 tickets that will win and the other 50 will lose. For all we know, all 100 could win or all 100 could lose. And that is probability.

In your example, the probability of winning is 1/2. That I agree. But we cannot arrive at that conclusion from the question.
Senior Manager
Joined: 23 Jun 2006
Posts: 387

### Show Tags

08 Aug 2007, 12:48
i was invited to try give definite answer (someone thinks that i know something you don't...)

anyway - correct answer is E (3/4)
D is wrong... indeed, D was correct if there were 50 winning tickets and 50 losing tickets, but as the question written this is not the case.

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Non-Human User
Joined: 09 Sep 2013
Posts: 9162
Re: Out of 100 lottery tickets, each having 50% chance of  [#permalink]

### Show Tags

07 Aug 2018, 06:44
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.
_________________
Re: Out of 100 lottery tickets, each having 50% chance of &nbs [#permalink] 07 Aug 2018, 06:44
Display posts from previous: Sort by