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

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

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New post 06 Aug 2007, 09:15
00:00
A
B
C
D
E

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Question Stats:

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

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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

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New post 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

answer = 3/4 (choice E)

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....
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New post 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.
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New post 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!
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New post 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.
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New post 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
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New post 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.
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New post 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.
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New post 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 ? :-D
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New post 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.
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New post 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.
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New post 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.
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New post 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.

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Re: Out of 100 lottery tickets, each having 50% chance of  [#permalink]

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