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Probability - 10 marbles

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by asax » Sat Jun 16, 2012 4:34 am
UFF.. after 2 days of failing at BTG daily qs. i finally got one right!
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by Brent@GMATPrepNow » Sat Jun 16, 2012 9:20 am
asax wrote:UFF.. after 2 days of failing at BTG daily qs. i finally got one right!
Great work, asax!

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by jcnasia » Mon Jun 25, 2012 6:58 am
bluementor wrote:
I don't get it. Why do you assume that all the girls will pick the marbles first before the boys do or vice versa? If the girls and boys picked a marble each in turns, the result would be different.

Could anyone please explain? Thanks.

-BM-
It doesn't matter what order they pick the marbles in. The answer is the same.

Prob(First girl picks any marble) = 1
Prob(First boy picks opposite-color marble) = 5/9
Prob(2nd girl picks same-color marble) = 4/8
Prob(2nd boy picks opp-color marble) = 4/7
Prob(3rd girl picks same-color marble) = 3/6
Prob(3rd boy picks opp-color marble) = 3/5
Prob(4th girl picks same-color marble) = 2/4
Prob(4th boy picks opp-color marble) = 2/3
Prob(5th girl picks same-color marble) = 1/2
Prob(5th boy picks opp-color marble) = 1

1*5/9*4/8*4/7*3/6*3/5*2/4*2/3*1/2*1= 5!4!/9! = (4*3*2) / (9*8*7*6) = 1/126
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by surmilsehgal » Fri Jul 06, 2012 11:15 pm
1/24
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by mba404 » Sat Aug 04, 2012 10:30 am
A

I approached it slightly differently. I figured out it is same as distributing these 1o balls (5W & 5B) to 10 people (5G & 5B).

Number of ways balls can be distributed so that all Girls get same colour = 2 (All white or All black)
Total number of ways of distributing = (10!)/(5!)(5!) = 512 (10! for distributing 10 different balls. As arrangements between same colored balls does not count in this case - divide 10! by 5!&5!)

probability = 2/512 = 1/126 (A)
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by tisrar02 » Sat Aug 04, 2012 12:00 pm
The question doesn't really care about the boys here so I only looked at the girls and came up with this:

1 girl chooses black marble
2nd chooses black marble
.
.
.
5th girl chooses black marble

Probability= 10/10*4/9*3/8*2/7*1/6= 24/3024= 1/126

A
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by tisrar02 » Sat Aug 04, 2012 12:00 pm
Double post.. Sorry Computer at work is messed up!
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by bjc7227 » Thu Sep 20, 2012 4:48 am
Brent@GMATPrepNow wrote:The answer is, indeed, A (1/126)
Here's my solution:
The easiest/fastest way to determine the probability is to examine the probability of each necessary outcome to guarantee that the girls (and subsequently the boys) draw the same colored marble.
We get [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(boys getting same color ball from remaining balls)
This equals: 1 x 4/9 x 3/8 x 2/7 x 1/6 x 1
Which equals: 1/126
Isn't there an issue with this question though. Our solutions generally assume that all of the girls necessarily choose first even though this is not explicitly stated in the problem. Wouldn't the answer change if we assumed that the girls and boys chose their marbles alternatively?
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by Brent@GMATPrepNow » Thu Sep 20, 2012 8:33 am
bjc7227 wrote:
Brent@GMATPrepNow wrote:The answer is, indeed, A (1/126)
Here's my solution:
The easiest/fastest way to determine the probability is to examine the probability of each necessary outcome to guarantee that the girls (and subsequently the boys) draw the same colored marble.
We get [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(boys getting same color ball from remaining balls)
This equals: 1 x 4/9 x 3/8 x 2/7 x 1/6 x 1
Which equals: 1/126
Isn't there an issue with this question though. Our solutions generally assume that all of the girls necessarily choose first even though this is not explicitly stated in the problem. Wouldn't the answer change if we assumed that the girls and boys chose their marbles alternatively?
The order makes no difference. To demonstrate this, I'll refer you to jcnasia's post (6 posts before this one)

jcnasia's approach
Prob(First girl picks any marble) = 1
Prob(First boy picks opposite-color marble) = 5/9
Prob(2nd girl picks same-color marble) = 4/8
Prob(2nd boy picks opp-color marble) = 4/7
Prob(3rd girl picks same-color marble) = 3/6
Prob(3rd boy picks opp-color marble) = 3/5
Prob(4th girl picks same-color marble) = 2/4
Prob(4th boy picks opp-color marble) = 2/3
Prob(5th girl picks same-color marble) = 1/2
Prob(5th boy picks opp-color marble) = 1

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

Other orders will yield the same answer.

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Brent
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by smanstar » Tue Sep 25, 2012 10:29 pm
What I did was
P( white marble) = 1/2
P ( black marble) = 1/2

So 5 girls selecting 5 white marbles and 5 boys selecting 5 black marbles
P = 1/2 * 1/2 * ...... 10 times = 1/2^10


Can someone explain where I am going wrong ???
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by Brent@GMATPrepNow » Wed Sep 26, 2012 6:07 am
smanstar wrote:What I did was
P( white marble) = 1/2
P ( black marble) = 1/2

So 5 girls selecting 5 white marbles and 5 boys selecting 5 black marbles
P = 1/2 * 1/2 * ...... 10 times = 1/2^10


Can someone explain where I am going wrong ???
This solution would be correct if each selected marbles were replaced after each selection. That way, the probability would be 1/2 that a child chooses the correct color marble.
However, if the marbles are not replaced, then the probabilities change.
For example, if a girl chooses first and she selects a white marble, then if another girl chooses next, there are now 9 marbles and 4 of them are white. So, the probability that she also chooses a white marble is 4/9. . . . etc.

I hope that helps.

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by mparakala » Fri Oct 19, 2012 6:17 am
Ans: 1/126 (A)

This is how I solved it. Please let me know if I am correct!

5 girls can choose one set of same colore marbles in 5C5 ways or another set of 5 colored marbles in 5C5 ways
=> 5C5 +5C5 = 1+ 1= 2

now, the entire set of 10 marbles can be chosen by 5 girls in 10C5 = 10!/ 5!5! = 252

P(of choosing the same set of 5 marbles out of a total 10) = 2/252 = 1/126
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by Brent@GMATPrepNow » Fri Oct 19, 2012 7:11 am
mparakala wrote:Ans: 1/126 (A)

This is how I solved it. Please let me know if I am correct!

5 girls can choose one set of same colore marbles in 5C5 ways or another set of 5 colored marbles in 5C5 ways
=> 5C5 +5C5 = 1+ 1= 2

now, the entire set of 10 marbles can be chosen by 5 girls in 10C5 = 10!/ 5!5! = 252

P(of choosing the same set of 5 marbles out of a total 10) = 2/252 = 1/126
Perfect!

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by sid128 » Thu Nov 15, 2012 1:07 am
I also got the same answer i.e. A (1/126) by applying following logic:

Total number of possibilities of distributing 5 white (identical) and 5 black (identical) marbles in 5 girls and 5 boys is (10!)/(5!)(5!)= 252

Favourable Events = 2 (Girls getting either all marbles or all black marbles)

Probability = No. of favourable events/No. of total events
= 2/252
= 1/126
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by Brent@GMATPrepNow » Thu Nov 15, 2012 9:10 am
sid128 wrote:I also got the same answer i.e. A (1/126) by applying following logic:

Total number of possibilities of distributing 5 white (identical) and 5 black (identical) marbles in 5 girls and 5 boys is (10!)/(5!)(5!)= 252

Favourable Events = 2 (Girls getting either all marbles or all black marbles)

Probability = No. of favourable events/No. of total events
= 2/252
= 1/126
Beautiful!
That's the great thing about probability questions - there are often many different possible approaches.

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Brent
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