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Re: Probability question, two different approaches? [#permalink]
1
Monco wrote:
A garden center has seven female and eight male holly plants, but they aren’t labeled as such. If a customer randomly purchases three plants, what is the probability of getting both male and female plants (thus holly berries, eventually)?


Write your answer as a reduced fraction.


P(getting both male and female plants)

There are 6 different ways to get both male and female plants when we randomly choose 3 plants:
1) MMF (male on 1st pick, male on 2nd pick, and female on 3rd pick)
2) MFM
3) FMM
4) FFM
5) FMF
6) MFF

So, P(getting both male and female plants) = P(MMF) + P(MFM) + P(FMM) + P(FFM) + P(FMF) + P(MFF)
= (8/15)(7/14)(7/13) + (8/15)(7/14)(7/13) + (7/15)(8/14)(7/13) + (7/15)(6/14)(8/13) + (7/15)(8/14)(6/13) + (8/15)(7/14)(6/13)
= 4/5

Cheers,
Brent
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Re: Probability question, two different approaches? [#permalink]
1
GreenlightTestPrep wrote:
Monco wrote:
A garden center has seven female and eight male holly plants, but they aren’t labeled as such. If a customer randomly purchases three plants, what is the probability of getting both male and female plants (thus holly berries, eventually)?


Write your answer as a reduced fraction.


P(getting both male and female plants)



There are 6 different ways to get both male and female plants when we randomly choose 3 plants:
1) MMF (male on 1st pick, male on 2nd pick, and female on 3rd pick)
2) MFM
3) FMM
4) FFM
5) FMF
6) MFF

So, P(getting both male and female plants) = P(MMF) + P(MFM) + P(FMM) + P(FFM) + P(FMF) + P(MFF)
= (8/15)(7/14)(7/13) + (8/15)(7/14)(7/13) + (7/15)(8/14)(7/13) + (7/15)(6/14)(8/13) + (7/15)(8/14)(6/13) + (8/15)(7/14)(6/13)
= 4/5

Cheers,
Brent



Wouldn't ( 7C1*8C2 + 8C1*7C2 )/15C3 , suffice...
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Re: Probability question, two different approaches? [#permalink]
Easier:

(7C2*8c1 + 8C2*7c1) / 15C3
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Re: Probability question, two different approaches? [#permalink]
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Re: Probability question, two different approaches? [#permalink]
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