Probability - 10 marbles

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by eaakbari » Thu Nov 15, 2012 11:44 am
Brent,

I computed as below.

1 x 4/9 x 3/8 x 2/7 x 1/6 x 1

But multiplied the above by 2 as there are 2 cases, one with black & one with white ball.

Where's my logic faltering?
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by Brent@GMATPrepNow » Thu Nov 15, 2012 11:51 am
eaakbari wrote:Brent,

I computed as below.

1 x 4/9 x 3/8 x 2/7 x 1/6 x 1

But multiplied the above by 2 as there are 2 cases, one with black & one with white ball.

Where's my logic faltering?
Presumably, you multiplied by two since it could be the case that the girls get all white marbles or all black marbles. However, your calculations already account for both possibilities.

Your calculations are: 1 x 4/9 x 3/8 x 2/7 x 1/6 x 1

The first 1 (I'm assuming) is the probability that the first person selects any color (white or black)
The next number (4/9) is the probability that the second person selects the same color as the first person selected(white or black)
And so on.

Since you've already accounted for both possibilities (girls get all white or girls get all black), there's no reason to multiply your earlier result by 2

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by sri_r » Thu Nov 29, 2012 12:53 am
@Brent - I have a doubt. I used a different logic and arrive at the solution. Could you please tell me if I am correct:

Probability of all girls selecting same marble => Probability of all girls selecting white marble or Probability of all girls black marble
=>Probability of all girls selecting white marble + Probability of all girls black marble

Probability of all girls selecting white marble = No. of ways of all girls selecting white marble/Total no. of ways of selecting white marble

= 5C5/10C5 = 1/10C5

Similar for Probability of all girls selecting black marble = 1/10C5

Therefore, Probability of all girls selecting same marble = 1/10C5 + 1/10C5 = 2/10C5 = 1/126

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by Brent@GMATPrepNow » Thu Nov 29, 2012 7:09 am
sri_r wrote:@Brent - I have a doubt. I used a different logic and arrive at the solution. Could you please tell me if I am correct:

Probability of all girls selecting same marble => Probability of all girls selecting white marble or Probability of all girls black marble
=>Probability of all girls selecting white marble + Probability of all girls black marble

Probability of all girls selecting white marble = No. of ways of all girls selecting white marble/Total no. of ways of selecting white marble

= 5C5/10C5 = 1/10C5

Similar for Probability of all girls selecting black marble = 1/10C5

Therefore, Probability of all girls selecting same marble = 1/10C5 + 1/10C5 = 2/10C5 = 1/126
Perfect - nice work!

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by davidschneider » Wed Jan 02, 2013 7:22 am
1/126

Great one!

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by bnpetteway » Tue Jan 08, 2013 5:31 pm
@Brent, I did it the same except why wouldn't you start the equation with (5/10) rather than 1 or (4/9)? If the question doesn't make sense then I will try m best to reword it.

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by Brent@GMATPrepNow » Tue Jan 08, 2013 5:41 pm
bnpetteway wrote:@Brent, I did it the same except why wouldn't you start the equation with (5/10) rather than 1 or (4/9)? If the question doesn't make sense then I will try m best to reword it.
I rewrote the probability as:
P(1st girls selects any marble AND 2nd girl selects marble the same color as 1st girl AND 3rd girl selects marble the same color as 1st girl AND 4th girl selects marble the same color as 1st girl AND 5th girl selects marble the same color as 1st girl AND the boys get the rest)
This equals P(1st girl selects any marble) x P(2nd girl selects marble the same color as 1st girl) x P(3nd girl selects marble the same color as 1st girl) x P(4th girl selects marble the same color as 1st girl) x P(5th girl selects marble the same color as 1st girl) x P(the boys get the rest)
This equals: 1 x 4/9 x 3/8 x 2/7 x 1/6 x 1
Which equals: 1/126

The 1 at the beginning of the product represents the probability that the 1st girls selects any marble. Since this first girl can select any marble, the probability is 1 that she will accomplish this.

I hope that helps.

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by bnpetteway » Tue Jan 08, 2013 5:47 pm
On the marble question: I did everything except for multiply 1 at the beginning and at the end. The 1 at the beginning, does it substitute for (5/10) and the 1 at the end, if it's the boys picks of marbles do we need it since it only asks about the girls?

Brent, what did you use to prepare for the GMAT? You and GuruNY are answering some pretty tough questions and your logic makes a lot of sense to me after you explain it but I would like to get the question when I first see it rather than someone explaining it to me. I really need a 700+ so hopefully you can fill me in.

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by Brent@GMATPrepNow » Tue Jan 08, 2013 6:03 pm
bnpetteway wrote:Brent, what did you use to prepare for the GMAT?
My answer probably won't help much since I had a ridiculously long prep time.
First, I taught many many GMAT prep courses over the course of several years.
Then, in order to get hired by a different test prep company, I had to get a certain score.
That's when I finally took the test.
So, basically, my preparation involved thousands of hours preparing for and teaching GMAT prep courses. It didn't hurt to have a math degree either :-)

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by bnpetteway » Tue Jan 08, 2013 6:30 pm
@Brent, can you post questions everyday, challenging questions because I need a lot of preparation to be able to make my 700+ score and I will answer to the best of my ability. Also, questions that will require a lot of thinking and logic as well.

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by Brent@GMATPrepNow » Tue Jan 08, 2013 6:35 pm
It's a lot of work to create challenging, GMAT-quality questions.

That said, I'd say that at least one-quarter of all questions posted on BTG fall into the 700+ category.

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by rajeshsinghgmat » Sat Jan 12, 2013 6:19 pm
(5*4*3*2*1)*(5*4*3*2*1)/(10*9*8*7*6*5*4*3*2*1)+(5*4*3*2*1)*(5*4*3*2*1)/(10*9*8*7*6*5*4*3*2*1)

=(2)*(5*4*3*2*1)*(5*4*3*2*1)/(10*9*8*7*6*5*4*3*2*1)
=(2)*(5*4*3*2*1)/(10*9*8*7*6)
=(4*1)/(9*8*7)
=1/(9*2*7)
=1/(63*2)
=1/126

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by darthy734 » Mon Mar 25, 2013 10:10 am
My Logic for answering this question:

Probability of the occurrence of an event = (Successful Attempts/Total no. of Attempts)

1. Successful Attempts = 2
(All the Girls pick up White marbles or All the Girls pick up Black marbles)

2. Total no. of Attempts = 2+2(5.5)+2(10.10) = 256

How is that ?

2 = Corresponds to the successful attempts

2(5.5) = Corresponds to WBBBB (5C1 ways) to Girls and BWWWW (5C1) to Guys, Multiplied it with 2 because it could as well be BWWWW to Girls and WBBBB to Guys.

2(10.10) = Corresponds to WWBBB (5C2 ways) to Girls and BBWWW (5C2) to Guys, Multiplied it with 2 because it could as well be BBWWW to Girls and WWBBB to Guys.

So P = 2/256 = 1/126 :)

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by Brent@GMATPrepNow » Mon Mar 25, 2013 2:49 pm
darthy734 wrote:My Logic for answering this question:

Probability of the occurrence of an event = (Successful Attempts/Total no. of Attempts)

1. Successful Attempts = 2
(All the Girls pick up White marbles or All the Girls pick up Black marbles)

2. Total no. of Attempts = 2+2(5.5)+2(10.10) = 256

How is that ?

2 = Corresponds to the successful attempts

2(5.5) = Corresponds to WBBBB (5C1 ways) to Girls and BWWWW (5C1) to Guys, Multiplied it with 2 because it could as well be BWWWW to Girls and WBBBB to Guys.

2(10.10) = Corresponds to WWBBB (5C2 ways) to Girls and BBWWW (5C2) to Guys, Multiplied it with 2 because it could as well be BBWWW to Girls and WWBBB to Guys.

So P = 2/256 = 1/126 :)
Hmmm, I don't quite follow the 5.5 and 10.10 parts, but the answer is correct.
So, I'm inclined to believe that your solution is good. :-)

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by smvjkumar » Sat Apr 06, 2013 11:04 am
first girl can pick any color so probability of that girl is 1
second girl have to pick the same color as first so probability = 4/ 9
the next one is 3/8, 2/7, 1/6

so total propability = 1X 4/9 X 3/8 X 2/7 X 1/6

= 1/126
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