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# At a certain riding school, students are randomly assigned

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Manager
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At a certain riding school, students are randomly assigned [#permalink]

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13 Oct 2004, 23:13
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At a certain riding school, students are randomly assigned to the horses they will use during any given lesson. The school has exactly 5 male horses and 6 female horses. If there are exactly 3 women and 8 men in the class, what is the probability that a student will be assigned a horse of the same sex?

I'll post the answer but this problem threw me off even though it's not that hard.
Manager
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18 Oct 2004, 16:44
OA is 58/121.

Possible assignment of 11 horses to 11 people = 121.
Possible ways to assign 6 female horses to 3 women = 18.
Possible ways to assign 5 male horses to 8 men = 40.
Possible ways to assign horse to someone of same gender = 18 + 48 = 58

Probability of being assigned to a horse of the same gender 58/121
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20 Oct 2004, 22:20
I got the same answer, slightly different way to look at it

P(Man) = 8/11 , P (Woman) = 3/11
P(Male Horse) = 5/11, P (Fem Horse) = 6/11

so the required prob = P(Man)*P(Male Horse) + P(Woman)*P(Fem Horse)
= 8/11*5/11 + 3/11*6/11 = (40 + 18)/121 = 58/121
Senior Manager
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24 Oct 2004, 02:45
PracticeMore wrote:
I got the same answer, slightly different way to look at it

P(Man) = 8/11 , P (Woman) = 3/11
P(Male Horse) = 5/11, P (Fem Horse) = 6/11

so the required prob = P(Man)*P(Male Horse) + P(Woman)*P(Fem Horse)
= 8/11*5/11 + 3/11*6/11 = (40 + 18)/121 = 58/121

wonderfull method buddy

thanks,
Dharmin
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Perseverance, Hard Work and self confidence

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24 Oct 2004, 18:13
nice... good method, i prefer in this way

==> # of possibles outcomes:

we need a couple of one horse (11C1) and one student (11C1), so we obtain: 11C1 * 11C1 = 11*11 = 121 couples

==> # of favorable outcomes:

A couple of one male horse (5C1) and one man (8C1), so we obtain:
5C1 * 8C1 = 5*8 = 40 couples

A couple of one fem horse (6C1) and one woman (3C1), so we obtain:
6C1 * 3C1 = 6*3 = 18 couples

==> Therefore, the probability will be: (40+18)/121 = 58/121

hope it helps
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24 Oct 2004, 21:48
I too used the method used by "PracticeMore". I guess, if one has a consistent method that works for all problems, then he/she is ok
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