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tkkishore
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Post subject: probability strategy questions 1-3  Posted: Wed Nov 28, 2007 11:47 pm |
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What is the probability that the sum of two dice will yield a 4 or 6 ?
When you say there are 36 ways the 2 dice are thrown, then shouldnt you count (1,3), (2,2), (2,2), (3,1) as all possibilities for sum of 4 ? Therefore, i contest that the solution to problem 1 should be (4 + 6)/36 or 5/18. The 6 comes from 6 possible combinations that can produce sum of 6.
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StaceyKoprince
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Post subject: Posted: Fri Nov 30, 2007 1:26 am |
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Nope - that's a common error. When you calculate possible outcomes for probability, you always do one step at a time: first one die, then the other. Don't try to do these as a pair of dice rolled simultaneously. When there are multiple conditions, you always set one condition first, because that first condition then determines the possibilities for the second condition.
If you roll a 1 first, then you need the second to be a 3 or a 5. That equals two possibilities.
If you roll a 2 first, then you need the second to be a 2 or a 4. That equals two possibilities.
If you roll a 3 first, then you need the second to be a 1 or a 3. That equals two possibilities.
If you roll a 4 first, then you need the second to be a 2. That equals one possibility.
If you roll a 5 first, then you need the second to be a 1. That equals one possibility.
If you roll a 6 first, you're stuck - zero possibilities.
_________________
Stacey Koprince
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Director of Online Community
ManhattanGMAT
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Amit05
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Post subject: Re: probability strategy questions 1-3 Posted: Tue Feb 12, 2008 12:40 am |
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tkkishore wrote: What is the probability that the sum of two dice will yield a 4 or 6 ?
When you say there are 36 ways the 2 dice are thrown, then shouldnt you count (1,3), (2,2), (2,2), (3,1) as all possibilities for sum of 4 ? Therefore, i contest that the solution to problem 1 should be (4 + 6)/36 or 5/18. The 6 comes from 6 possible combinations that can produce sum of 6.
So, is the answer 2/9 ?
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shaji
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Post subject: Re: probability strategy questions 1-3 Posted: Tue Feb 12, 2008 7:14 am |
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Amit05 wrote: tkkishore wrote: What is the probability that the sum of two dice will yield a 4 or 6 ?
When you say there are 36 ways the 2 dice are thrown, then shouldnt you count (1,3), (2,2), (2,2), (3,1) as all possibilities for sum of 4 ? Therefore, i contest that the solution to problem 1 should be (4 + 6)/36 or 5/18. The 6 comes from 6 possible combinations that can produce sum of 6. So, is the answer 2/9 ? Precisely!!!
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StaceyKoprince
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Post subject: Posted: Fri Feb 15, 2008 1:09 am |
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Yep, you've got it, guys!
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Stacey Koprince
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ManhattanGMAT
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melissa_curran
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Post subject: Re: probability strategy questions 1-3 Posted: Thu Apr 15, 2010 4:01 pm |
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Hi, I don't understand why you look and rolling the dice simulataneously which would make the answer 5/18 on question 1 (incorrect). Can you elaborate on why you look at rolling one at a time?
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Ben Ku
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Post subject: Re: probability strategy questions 1-3 Posted: Sat May 01, 2010 2:12 am |
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melissa_curran wrote: Hi, I don't understand why you look and rolling the dice simulataneously which would make the answer 5/18 on question 1 (incorrect). Can you elaborate on why you look at rolling one at a time? Hi Melissa, It's not really about rolling them one at a time; it's more about there being two dice. Imagine if one die is red and the other is white. Rolling a red 1 white 3 is different from rolling a red 3 and white 1. Hope that makes sense.
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Ben Ku
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ManhattanGMAT
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pellucide
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Post subject: Re: probability strategy questions 1-3 Posted: Sun Dec 19, 2010 8:29 pm |
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tkkishore wrote: What is the probability that the sum of two dice will yield a 4 or 6 ?
When you say there are 36 ways the 2 dice are thrown, then shouldnt you count (1,3), (2,2), (2,2), (3,1) as all possibilities for sum of 4 ? Therefore, i contest that the solution to problem 1 should be (4 + 6)/36 or 5/18. The 6 comes from 6 possible combinations that can produce sum of 6. I dont think you can count (2,2) twice. Also you cannot count (3,3) twice. So with the way you describe, the favourable outcomes should be (1,3), (2,2), (3,1) - 3 outcomes for a 4-total (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes for a 6-total So total favourable outcomes = 3+5 Total number of outcomes = 36 Probability = 8/36 = 2/9
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jnelson0612
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Post subject: Re: probability strategy questions 1-3 Posted: Mon Dec 20, 2010 9:16 am |
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Correct, pellucide. Nice work!
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Jamie Nelson
ManhattanGMAT Instructor
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coreygillard
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Post subject: Re: probability strategy questions 1-3 Posted: Sat Mar 05, 2011 6:07 pm |
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Why can't you count (2,2) twice? You count (1,3) & (3,1)...isn't this the same principal? One die rolls a 2, the other rolls a 2. I'm confused why it works with 3 and 1, but not 2 & 2.
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tim
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Post subject: Re: probability strategy questions 1-3 Posted: Sun Mar 06, 2011 10:57 pm |
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imagine you have two different colored dice. rolling a blue 1 and a red 3 is different from rolling a red 1 and a blue 3. rolling a blue 2 and a red 2 is the same as rolling a red 2 and a blue 2..
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Tim Sanders
Manhattan GMAT Instructor
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