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Probability: where am I going wrong?

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Probability: where am I going wrong?

by loveusonu » Thu Aug 26, 2010 1:26 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 1/7 , what is the probability that either Mike or Rob will win the championship?(there can be only one winner)

a. 1/12
b. 1/7
c. 1/2
d. 7/12
e. 6/7


My solution:

p(ben winning) = 1/7 , therefore p(ben loosing) = 1 - 1/7 =6/7
p(either mike or rob wins) = 6/7 * [ 1/4 * (1 - 1/3) + 1/3 * (1- 1/4)] = 7/12

The OA says, p(either mike or rob wins) = 6/7 * [ 1/4 + 1/3] = 1/2 :(

Am I missing some logic here?
Sonu
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by phillybeat » Thu Aug 26, 2010 1:35 pm
loveusonu 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 1/7 , what is the probability that either Mike or Rob will win the championship?(there can be only one winner)

a. 1/12
b. 1/7
c. 1/2
d. 7/12
e. 6/7


My solution:

p(ben winning) = 1/7 , therefore p(ben loosing) = 1 - 1/7 =6/7
p(either mike or rob wins) = 6/7 * [ 1/4 * (1 - 1/3) + 1/3 * (1- 1/4)] = 7/12

The OA says, p(either mike or rob wins) = 6/7 * [ 1/4 + 1/3] = 1/2 :(

Am I missing some logic here?

P(mike) = 6/7*1/4
p(rob) = 6/7*1/3


since you got the probability of Mike or Rob winning from the Question , you just use those..


either one of them winner = 6/7(1/4+1/3) = 6/6*7/12 = 1/2

hope it helps..
Last edited by phillybeat on Thu Aug 26, 2010 1:38 pm, edited 1 time in total.
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by beatthegmatinsept » Thu Aug 26, 2010 1:35 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 1/7 , what is the probability that either Mike or Rob will win the championship?(there can be only one winner)

a. 1/12
b. 1/7
c. 1/2
d. 7/12
e. 6/7


My solution:

p(ben winning) = 1/7 , therefore p(ben loosing) = 1 - 1/7 =6/7 Correct

p(either mike or rob wins) = 6/7 * [ 1/4 * (1 - 1/3) + 1/3 * (1- 1/4)] = 7/12

The OA says, p(either mike or rob wins) = 6/7 * [ 1/4 + 1/3] = 1/2 :(

Am I missing some logic here?

I would calculate Mike and Rob's probability separately,

So Mike wins, P = 6/7 * 1/4 = 3/14
Rob wins, P = 6/7 * 1/3 = 2/7.

So P that EITHER Mike or Rob win, 3/14 + 2/7 = 1/2.
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by shaw3257 » Thu Aug 26, 2010 1:55 pm
beatthegmatinsept 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 1/7 , what is the probability that either Mike or Rob will win the championship?(there can be only one winner)

a. 1/12
b. 1/7
c. 1/2
d. 7/12
e. 6/7
Probability of Ben losing the championship: 1 - 1/7 = 6/7
Probability of Mike Winning the championship: 1/4
Probability of Rob Winning the championship: 1/3

Probability of Rob or Mike winning: P(B Losing) * P(R or M) = P(B Losing) * ( P(R) + P(M) )

Probability of Rob or Mike winning: (6/7) * (1/4 + 1/3) = 1/2
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by Brian@VeritasPrep » Thu Aug 26, 2010 2:24 pm
Hey, loveusonu:

Great question! Just reading what you wrote, I think you're complicating this one a little more than you need to. In your solution, it looks like you're taking:

6/7 chance that Ben loses (correct)
1/4 chance that Mike wins (correct), but only in the 2/3 (you have it as 1 -1/3) chance that Rob loses (that's where you're incorrect)

The question says that there can only be one winner, so you don't need to take out the 1/3 chance of Rob winning to calculate the probability of Mike winning, because they won't both win. The combined probability is 0.

Therefore, that extra step you take of subtracting the probability of "both" in your calculations for Mike and for Rob is unnecessary and, actually, incorrect.




Now, the reason that you have to do it for Ben is that we only know that Mike's probability is 1/4 "If Ben were to lose". It's not an overall 25% chance that he wins...it's a 25% chance if Ben were to lose.


I hope that helps...
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by beatthegmatinsept » Thu Aug 26, 2010 2:42 pm
@ Brian - What difficulty level do you think this question falls under?
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by Brian@VeritasPrep » Thu Aug 26, 2010 3:32 pm
Hey, beatthegmatinsept (and Sept is coming soon!):

Good question - I'm always a little uncomfortable giving difficulty estimates, but this is definitely above average...somewhere between 600 and 700, I'd say, but it's hard to narrow in too much more than that. It's tough - spending so much time on this board and with Veritas students, I work with pretty self-selecting groups of people so it's hard for me to gauge genuine difficulty, since so many of the people I talk to are significantly above average on the test.

Since you mentioned difficulty, let me add this blog post I did last week about the GMAT scoring algorithm: https://blog.veritasprep.com/2010/08/und ... rithm.html
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by beatthegmatinsept » Thu Aug 26, 2010 3:39 pm
Brian@VeritasPrep wrote:Hey, beatthegmatinsept (and Sept is coming soon!):

Good question - I'm always a little uncomfortable giving difficulty estimates, but this is definitely above average...somewhere between 600 and 700, I'd say, but it's hard to narrow in too much more than that. It's tough - spending so much time on this board and with Veritas students, I work with pretty self-selecting groups of people so it's hard for me to gauge genuine difficulty, since so many of the people I talk to are significantly above average on the test.

Since you mentioned difficulty, let me add this blog post I did last week about the GMAT scoring algorithm: https://blog.veritasprep.com/2010/08/und ... rithm.html
Thanks Brian. I hope I don't have to change my username from beatthegmatinsept to beatthegmatinoct ;) Keeping my fingers Crossed!
btw, Interesting Blog!
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by loveusonu » Fri Aug 27, 2010 3:25 am
Brian@VeritasPrep wrote:Hey, loveusonu:

The question says that there can only be one winner, so you don't need to take out the 1/3 chance of Rob winning to calculate the probability of Mike winning, because they won't both win. The combined probability is 0.

Therefore, that extra step you take of subtracting the probability of "both" in your calculations for Mike and for Rob is unnecessary and, actually, incorrect.
Thanks Brian,
This the exact concept I was looking for :)

I wish to drive deeper for this concept. If you have any links\question regarding this concept, please share.

Thanks again.
Sonu
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by Brian@VeritasPrep » Fri Aug 27, 2010 8:26 am
Glad that helped, Sonu!

I don't like being a corporate shill on here, so it's awkward for me to say, but let me put it in more of a personal recommendation form - Chris@VeritasPrep wrote the majority of our book on Probability and Combinatorics and it's incredibly good. If you're looking for GMAT probability help, that's the best resource I've seen. It's available at Amazon and now, I think, on the iPad, or through any of our GMAT programs.

I guess that's another way of saying that I'm spoiled - I don't know of too many links or outside resources for probability that I'd recommend because I've never had to go any farther than Chris' book.

Actually, I'll tell you what - I'm due to write up a "GMAT Tip of the Week" blog post for https://blog.veritasprep.com today. I'll write one about probability in your honor! Should be up in an hour or so...

Cheers,
Brian
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by loveusonu » Fri Aug 27, 2010 8:41 pm
Brian@VeritasPrep wrote:Glad that helped, Sonu!

I don't like being a corporate shill on here, so it's awkward for me to say, but let me put it in more of a personal recommendation form - Chris@VeritasPrep wrote the majority of our book on Probability and Combinatorics and it's incredibly good. If you're looking for GMAT probability help, that's the best resource I've seen. It's available at Amazon and now, I think, on the iPad, or through any of our GMAT programs.

I guess that's another way of saying that I'm spoiled - I don't know of too many links or outside resources for probability that I'd recommend because I've never had to go any farther than Chris' book.

Actually, I'll tell you what - I'm due to write up a "GMAT Tip of the Week" blog post for https://blog.veritasprep.com today. I'll write one about probability in your honor! Should be up in an hour or so...

Cheers,
Brian
Thanks Brian, That blog would be my notes.
Sonu
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