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Probability Problem

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Probability Problem

by gmatusa2010 » Thu Sep 02, 2010 8:46 pm
X has 50% chance of winning a game, Y has 40% of winning the game, their chances are independent, what is the chance of one them winning? Only one can win.


My reasoning: there are 3 players in this game X, Y, and Z where Z= the OTHERS could be 1 or more doesn't matter. Z has 10% chance of winning so 1-10% = 90%; this is however wrong. Can somebody explain, shouldn't all the chances add up to 1. Afterall, one of the player has to win.
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by neerajkumar1_1 » Thu Sep 02, 2010 9:11 pm
gmatusa2010 wrote:X has 50% chance of winning a game, Y has 40% of winning the game, their chances are independent, what is the chance of one them winning? Only one can win.


My reasoning: there are 3 players in this game X, Y, and Z where Z= the OTHERS could be 1 or more doesn't matter. Z has 10% chance of winning so 1-10% = 90%; this is however wrong. Can somebody explain, shouldn't all the chances add up to 1. Afterall, one of the player has to win.
I suppose u got the problem...
the chances of x and y winning are independent... they have to relation... and there is certainly no Z...

what u need to do is...
chances of one of them winning would be:
either x wins and y loses or y wins and x loses...

therefore the required probability will be:
.5*.6 + .4*.5

.3+.2 = .5


Hope this helps...
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by Rahul@gurome » Thu Sep 02, 2010 9:13 pm
Solution:
There can be two cases.
Probability of X losing is (100 - 50)% or 50% and probability of Y losing is (100 - 40)% or 60%.
Case (1) X wins and Y loses. This probability will be 50% *60% = 30% .
Case (2) Y wins and X loses. This probability will be 40% * 50% = 20% .
Or required probability is 30% + 20% = 50%
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by gmatusa2010 » Thu Sep 02, 2010 9:35 pm
Is it reasonable to think 1 person with 50% second with 40% and someone else or a group has 10%? Does the % have to add up to 100%?

Because according to your answer there are 3 people: 1 with 50% 2 with 40% and 3 (someone else) 50%.
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by gmatusa2010 » Thu Sep 02, 2010 9:43 pm
Here's the official answer, it could be wrong:


1) 1-50%= 50% of not winning
2) 1-40%= 60% of not winning
3) Chance that neither win? 30%
4) Chance that at least one of them win? 1-30% = 70%
5) OA = 70%
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by neerajkumar1_1 » Thu Sep 02, 2010 9:45 pm
gmatusa2010 wrote:Is it reasonable to think 1 person with 50% second with 40% and someone else or a group has 10%? Does the % have to add up to 100%?

Because according to your answer there are 3 people: 1 with 50% 2 with 40% and 3 (someone else) 50%.
oh no no no.... u r getting it again wrong...

there are no 3 people here...

the probabilities are given for individual person winning or losing...

the probabilities will add up to one for each of them ...

so the prob of x winning and losing = 1...

hope that clarifies ur doubt..
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by gmatusa2010 » Thu Sep 02, 2010 9:52 pm
could this problem be flawed? how can the % actually be independent from each other if one of them has to win. Here's another way to ask this question. Can you explain how the original question is different to this question

"In a lottery, there are 10 tickets. A gets 5 tickets, B get 4 tickets, and C get 1 ticket, what's the chance of A or B winning?"
Here there's a finite number of tickets. How is it possible that you have more than 10 tickets? There's 1 draw.

I kinda don't understand the explanation above because once the first person win the other person not winning is 100%
To me this doesn't make sense : X wins Y doesn't win ==> 50% x 60%. That doesn't exist. What is the % that X wins? 50%. X's winning means Y loses automatically. The case that neither win is actually realistic----> meaning multiplying two NEITHER WIN % together makes sense because it can actually happen and the % doesn't change. WHere the 70% make sense is when this occurs 1) both win and the game is played over again until one 1 win. 2) % that both wins? 50%*40%= 20% 3) since 90% that one or both of them win but the winning is conditioned on only one of them win, you actually subtract the chance they both win out so 90-20- 70%. To me that's what the OA is implying but thats an entirely different game from what is described in the question. so poor question?


Can you make a game where the % of winning is independent but there's only one winner? That sounds unrealistic.
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by Ian Stewart » Thu Sep 02, 2010 10:56 pm
gmatusa2010 wrote:could this problem be flawed?
Deeply, deeply flawed. It's open to different interpretations, for one thing, and the solution that you pasted in from the source is wrong, at least if there can be only one winner.

While it isn't at all clear what the question means, I imagine they're thinking about a game that one player plays, then the other player plays; the two people are not playing against each other. So perhaps they're each doing something like playing chess against a computer, and X is a bit better than Y. Now, when the question says that 'only one can win', I can only guess that it's just trying to make clear that it's asking for the probability of exactly one of them winning (that's why their solution is wrong: they are including the possibility that X and Y both win). I don't think they're trying to say that it's impossible that both players win. Still, I interpreted it the same way as you did the first few times I read it, so the wording is terrible here. Indeed, if it is impossible for both people to win, then the probabilities aren't independent, so the question becomes mathematical nonsense.

Questions in Probability/Counting tend to require more careful wording than questions in any other GMAT area. Sources that can't be bothered to word their probability/counting questions precisely are going to be more confusing than helpful, and I'd recommend you not waste your time with them. Where is this question from?
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by gmatusa2010 » Thu Sep 02, 2010 11:09 pm
Nova's GMAT Bible by Derrick Vaughn and Jeff Kolby

I am trying different sources because I've exhausted MGMAT and OG. I just need questions to build up speed and practice time management. Do you recc any sources?
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