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Probability boys girls

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Probability boys girls

by El Cucu » Mon Mar 30, 2009 7:39 am
7 girls 3 boys 10 seats, probability of boys seating separately?

I am a bit lost.

It is ok to calculate the probability of seating together divided by the total outcomes ? Best aproach? tksvm
Last edited by El Cucu on Mon Mar 30, 2009 4:50 pm, edited 1 time in total.
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by moutar » Mon Mar 30, 2009 7:53 am
I'm not very good with probability but I would say...

Total number of ways of sitting 10 people = 10!
Number of ways boys would be sitting together:
BBBBBBBGGG
GBBBBBBBGG
GGBBBBBBBG
GGGBBBBBBB
=4

Prob boys seating separately = 1 - 4/10!

which seems a little steep.
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by El Cucu » Mon Mar 30, 2009 8:28 am
moutar wrote:I'm not very good with probability but I would say...

Total number of ways of sitting 10 people = 10!
Number of ways boys would be sitting together:
BBBBBBBGGG
GBBBBBBBGG
GGBBBBBBBG
GGGBBBBBBB
=4

Prob boys seating separately = 1 - 4/10!

which seems a little steep.
Answer is 14/15
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by moutar » Mon Mar 30, 2009 8:34 am
Let me try again.

Total number of ways of sitting 10 people = 10!
Number of ways boys would be sitting together:
BBBBBBBGGG (7! x 3!)
GBBBBBBBGG
GGBBBBBBBG
GGGBBBBBBB
= 7! x 3! x 4

Prob boys seated together = 7! x 3! x 4/10!
= 2 x 3 x 4/8 x 9 x 10
= 1/30

Prob boys sitting apart = 29/30

I hate probability.
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Re: Probability boys girls

by Vemuri » Mon Mar 30, 2009 10:00 am
How can only 3 girls separate 7 boys? Doesn't sound logical.
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by scoobydooby » Mon Mar 30, 2009 10:10 am
agree with Vemuri. 7 boys cant sit separately in 10 seats.

guess the question was about "no girls" being together?

consider the case where all girls are together and take them as a unit of 3 girls. so we have to arrange 7 boys and 1 unit (of 3 girls) or 8 items
in 8! ways. the girls themselves can be arranged in 3! ways within the unit.
so possible arrangements with girls being together 8!*3! ways

10 people can be arranged in 10 seats in 10! ways

cases in which no girls are together: 10!-8!*3!

probabilty of no girls being together: 10!-8!*3!/10!
=1-[(8!*3!)/10!]
=1-3/45
=42/45
=14/15
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by cm47323 » Mon Mar 30, 2009 12:38 pm
scoobydooby wrote: probabilty of no girls being together: 10!-8!*3!/10!
You solved P(girls not all together), different from P(no girls together). This question was poorly worded from the start...
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Re: Probability boys girls

by El Cucu » Mon Mar 30, 2009 4:48 pm
El Cucu wrote:7 boys 3 girls 10 seats, probability of boys seating separately?

I am a bit lost.

It is ok to calculate the probability of seating together divided by the total outcomes ? Best aproach? tksvm
Very sorry for the typo. Is 7 girls and 3 boys, probability of the boys seating separately. ( So I edited the original question)
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Re: Probability boys girls

by Vemuri » Mon Mar 30, 2009 5:48 pm
El Cucu wrote:
El Cucu wrote:7 boys 3 girls 10 seats, probability of boys seating separately?

I am a bit lost.

It is ok to calculate the probability of seating together divided by the total outcomes ? Best aproach? tksvm
Very sorry for the typo. Is 7 girls and 3 boys, probability of the boys seating separately. ( So I edited the original question)
Hi El, it would be helpful if you can provide the answer options as well.
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by gmat740 » Mon Mar 30, 2009 9:06 pm
I tried solving it but it looks to me that this question is not fit for GMAT considering the time crunch

Would love to see anybody answer this considering the time restriction

Karan
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by El Cucu » Tue Mar 31, 2009 5:19 pm
gmat740 wrote:I tried solving it but it looks to me that this question is not fit for GMAT considering the time crunch

Would love to see anybody answer this considering the time restriction

Karan
I owe an explanation to all who tried to solve the question.

Well, after 30 minutes of thinking, etc...I consider the problem:

[1- (together/total combinations)] So 1- 8/120= 14/15

Tks for your patience.
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