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Probability empty seats

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by moutar » Mon Mar 30, 2009 7:59 am
Total number of outcomes: 10!

Total number with 2 seats together: 9 x 8!
SSGGGGGGGG (8! ways of arranging the girls)
GSSGGGGGGG
GGSSGGGGGG
GGGSSGGGGG
GGGGSSGGGG
GGGGGSSGGG
GGGGGGSSGG
GGGGGGGSSG
GGGGGGGGSS

Prob = 9 x 8!/10! = 9/9 x 10 = 1/10

Any chance of options? I'm not very good with prob. This could be wrong.
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by El Cucu » Mon Mar 30, 2009 8:32 am
moutar wrote:Total number of outcomes: 10!

Total number with 2 seats together: 9 x 8!
SSGGGGGGGG (8! ways of arranging the girls)
GSSGGGGGGG
GGSSGGGGGG
GGGSSGGGGG
GGGGSSGGGG
GGGGGSSGGG
GGGGGGSSGG
GGGGGGGSSG
GGGGGGGGSS

Prob = 9 x 8!/10! = 9/9 x 10 = 1/10

Any chance of options? I'm not very good with prob. This could be wrong.
Answer is 1/5
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by shidoshide » Mon Mar 30, 2009 12:48 pm
El Cucu wrote:
moutar wrote:Total number of outcomes: 10!

Total number with 2 seats together: 9 x 8!
SSGGGGGGGG (8! ways of arranging the girls)
GSSGGGGGGG
GGSSGGGGGG
GGGSSGGGGG
GGGGSSGGGG
GGGGGSSGGG
GGGGGGSSGG
GGGGGGGSSG
GGGGGGGGSS

Prob = 9 x 8!/10! = 9/9 x 10 = 1/10

Any chance of options? I'm not very good with prob. This could be wrong.
Answer is 1/5
Because two empty seats are identical items, total number of outcomes should be 10! / 2.
That means the answer is 1/5.
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Re: Probability empty seats

by vittalgmat » Mon Mar 30, 2009 3:15 pm
El Cucu wrote:8 girls 10 seats probability of the 2 empty seats being together?

Total number of outcomes: 10C8 = 10C(10-8) = 10C2
= (10*9)/(2*1) = 45

Total number of pairs of empty seats: 9
ie. Assume that the seats are labelled 1 thru 10.
empty seat pairs can be written as below
(1,2), (2,3) .... (9,10) => there are 9 such pairs.

So probability = 9/45 = 1/5

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