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<need expert opinion> Probability question

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<need expert opinion> Probability question

by voodoo_child » Mon Oct 03, 2011 4:58 am
If Ben were to lose the championship, Mike would be the winner with a probability of 1/4 , and Rob 1/3 . If the probability of Ben being the winner is 1/7 , what is the probability that either Mike or Rob will win the championship?

(A) 1/12
(B) 1/7
(C) 1/2
(D) 7/12
(E) 6/7

OA C
Here's what i did...what's wrong here?
P(Mike winning)=P(Ben looses). P(Mike wins) . P(Rob looses)
P(Rob winning)=P(Ben looses). P(Mike looses) . P(Rob wins)

Adding the two will give us the answer right?

Now given,P(mike wins)=1/4.Therefore,P(mike looses)=3/4
Similarly,P(rob wins)=1/3.Therefore,P(rob looses)=2/3
And,P(ben wins)=1/7.Therefore,P(ben looses)=6/7

Putting the values,
6/7.1/4.2/3 + 6/7.3/4.1/3 = 1/7+3/14=5/14
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by sl750 » Mon Oct 03, 2011 6:01 am
The probability of Mike winning is only conditional on Ben losing. Ditto for Rob winning

P(Mike or Ben winning) = P(Ben losing)*(p(Mike winning) + p(Ben winning))

6/7(1/4+1/3) = 6/7*7/12 = 1/2
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by GmatMathPro » Mon Oct 03, 2011 7:31 am
Here's what i did...what's wrong here?
P(Mike winning)=P(Ben looses). P(Mike wins) . P(Rob looses)
P(Rob winning)=P(Ben looses). P(Mike looses) . P(Rob wins)

Adding the two will give us the answer right?

Now given,P(mike wins)=1/4.Therefore,P(mike looses)=3/4
Similarly,P(rob wins)=1/3.Therefore,P(rob looses)=2/3
And,P(ben wins)=1/7.Therefore,P(ben looses)=6/7

Putting the values,
6/7.1/4.2/3 + 6/7.3/4.1/3 = 1/7+3/14=5/14
The problem with this is that these probabilities are not independent. If mike wins, Rob's probability of losing is now 100%. So once you say "mike wins", rob's probability of losing shifts from 2/3 to 1. Same for the other one. Once you say "rob wins", mike's probability of losing goes from 3/4 to 1.
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by knight247 » Mon Oct 03, 2011 8:51 am
Pete,
Bro got some weird queries for you.

Now, Assuming that Ben loses which has a 6/7 chance of happening, there are two possibilities i.e. either Mike wins or Rob wins. So we have Ben Loses(Mike Wins+Rob Wins)=6/7(1/4 + 1/3)=1/2

What I have difficulty understanding is that if we assume that Ben loses and Mike wins, then why don't we have to multiply the probability of Rob losing??? Similarly if we consider Ben losing and Rob winning why don't he have to multiply by the probability of Mike losing.

In short, why is it not
Ben Loses(Mike Wins*Rob Loses+Rob Wins*Mike Loses) ???

I understand that they are mutually exclusive i.e. both can't happen simultaneously etc so no need to consider the 'losing' probability blah blah. If thats the case why consider the 6/7 possibility of Ben losing at all? Why not just take 1/4+1/3? I'm kinda confused with this. Hoping u could clarify. Thanks a ton.
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by shankar.ashwin » Mon Oct 03, 2011 9:22 am
I had the same doubt but as Pete clarifies,

When Ben loses, either Rob or Mike wins. If you consider one of them win, the probability of the other losing is sure (So its actually multiplied by 1)

So here it would be 6/7 ( 1/4*1 + 1/3*1)

1/4*1 - Mike wins, Rob loses
and, 1/3 * 1 - Rob wins, Mike loses.

Hope I am right here.
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by GmatMathPro » Mon Oct 03, 2011 9:36 am
Because those probabilities are entirely contingent upon Ben losing:

IF some event happens(Ben losing), THEN these probabilities become valid. If Ben doesn't lose, we don't know what their probabilities of winning are. All things being equal, Rob and Mike are both better off if Ben is out of the running. With Ben in the running, their probabilities are something unknown. If he is OUT of the running their chances are 1/4 and 1/3. But these are only valid if we know for sure that Ben doesn't win, so we have to include that probability in the calculation.
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by GmatMathPro » Mon Oct 03, 2011 9:39 am
Shankar,

Your math looks good to me, but note that your statement "when Ben loses, either Rob or Mike wins" isn't quite right. For one thing, 1/4+1/3 is not 1. For another, if that were true, then the probability of one of them winning would just be the same as the probability of Ben losing, 6/7. There must be some competitors that are never introduced to us.
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by gmatboost » Mon Oct 03, 2011 10:03 am
In general in any probability calculation, each subsequent probability that we introduce into the multiplication is conditional on the occurrence of all of the previous events.

When we say that the probability of flipping three consecutive heads is 1/2 * 1/2 * 1/2, we are really saying that:
p(3 heads) = p(1st head) * p(2nd head, given first flip was heads) * p(3rd head, given first flip was heads and second flip was heads)

When we say that an event is independent of all other events, we are more or less saying that we can ignore the conditional statements because they have no impact on the specific event.

In this case, as Pete said, the probability that Mike wins is 1/4 conditional on Bob losing. These events aren't independent. We are not told that Mike's chance of winning is 1/4 overall.

Note that we are making an assumption that is not explicitly stated. Namely, that exactly one person can win the championship. But I think that's a safe one (and one that would not be left unclear on the GMAT).
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by shankar.ashwin » Mon Oct 03, 2011 10:11 am
Ah! I was wondering why none of these numbers add up to 1. Usually we dont see questions like this where there are possibilities outside the scope of the question. But ya, I got what you said. Someone has to win and definitely there are more than 3 people here.
GmatMathPro wrote:Shankar,

Your math looks good to me, but note that your statement "when Ben loses, either Rob or Mike wins" isn't quite right. For one thing, 1/4+1/3 is not 1. For another, if that were true, then the probability of one of them winning would just be the same as the probability of Ben losing, 6/7. There must be some competitors that are never introduced to us.
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by voodoo_child » Tue Oct 04, 2011 3:29 am
Thanks Pete, Greg and BTG team for wonderful explanations. I didn't see the conditional probabilities.

Here's what I did: (Can you please confirm whether this solution looks fine?)

Set : X = Event that X wins (X= Mike(M), B or R)
X = Event that X loses (X = M, B or R)

1) P(M/B) = 1/4 ;
2) P(R/B) = 1/3 ;
3) Question : Find P(M Intersection B) + P(R Intersection B)
= P(M/B)*P(B) + P(R/B)*P(B)
= 1/4 * 6/7 + 1/3*6/7
= 1/2

Is this correct?
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by voodoo_child » Sun Oct 09, 2011 4:55 pm
Here's another thought from another forum : If we assume that there are 84 games and BEn loses in 6/7 of them => Ben loses in 72 games.

Now Mike has a probability of winning in 1/4 of the games => 18 games. Similarly, Rob in 24 games. My question is that who would win in the remaining 84-18-24-12 games = 30 games? Not sure though....

Any thoughts?
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by GmatMathPro » Sun Oct 09, 2011 5:41 pm
voodoo_child wrote:Here's another thought from another forum : If we assume that there are 84 games and BEn loses in 6/7 of them => Ben loses in 72 games.

Now Mike has a probability of winning in 1/4 of the games => 18 games. Similarly, Rob in 24 games. My question is that who would win in the remaining 84-18-24-12 games = 30 games? Not sure though....

Any thoughts?
Who wins those 30 games? Somebody besides Ben, Rob, and Mike whom we haven't had the pleasure of meeting. We're never told that they are the only competitors in this tournament. It's related to what I was saying to shankar. 1/4+1/3=7/12. So 5/12 of the time when ben loses, someone besides Rob and Mike wins the championship. 5/12 of 72 is 30, so that's where the "extra" 30 is coming from.
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