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Contidional Probability?

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by sumithshah » Sat Sep 27, 2008 11:13 pm
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 , what is the probability that either Mike or Rob will win the championship?

1/12 ; 1/7 ; 1/2 ; 7/12 ; 6/7

OA C


I have no clue how do this. Any pointers would be appreciated
Last edited by sumithshah on Sun Sep 28, 2008 12:24 am, edited 1 time in total.
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by manulath » Sat Sep 27, 2008 11:58 pm
please check the question, it is missing data - prababilities!
regards
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by sumithshah » Sun Sep 28, 2008 12:24 am
fixed
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by 4meonly » Sun Sep 28, 2008 2:37 am
not fixed yet

winner is ???????, what is the
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Re: Contidional Probability?

by manulath » Sun Sep 28, 2008 3:17 am
[quote="sumithshah"]If the probability of Ben being the winner is ,


Please correct the question. Data missing.
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Re: Contidional Probability?

by sudhir3127 » Sun Sep 28, 2008 3:27 am
sumithshah wrote: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 , what is the probability that either Mike or Rob will win the championship?

1/12 ; 1/7 ; 1/2 ; 7/12 ; 6/7

OA C


I have no clue how do this. Any pointers would be appreciated


This is the right Question

"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?" Sumit would appreciate if u double check before posting a Question .. it would unnecessarily delay the posters time ....


here goes the Answer...

Prob of Ben wins=1/7

Prob of ben loses =6/7.

P(Mike wins or Rob wins) = P(Mike wins)+P(Rob wins) - P(Both wins)
=1/4+1/3-0
=7/12.

But we are only 6/7 sure about Ben loosing.

hence the probability is = 6/7*7/12=1/2

hope that helps....
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by uptowngirl92 » Fri Sep 18, 2009 5:30 pm
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 :oops:

please explain where I am going wrong.
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by Stuart@KaplanGMAT » Fri Sep 18, 2009 8:23 pm
uptowngirl92 wrote: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 :oops:

please explain where I am going wrong.
You're double counting losses.

The only conditional part is Ben losing. There's no reason to multiply your individual events by the probability of Rob/Mike losing as well.

So, if you just want to add the two events:

Ben losing then Rob winning = 6/7 * 1/3 = 6/21 = 2/7

Ben losing then Mike winning = 6/7 * 1/4 = 6/28 = 3/14

2/7 + 3/14 = 4/14 + 3/14 = 7/14 = 1/2
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