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A certain basket contains 10 apples, 7 of which are red and

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A certain basket contains 10 apples, 7 of which are red and [#permalink] New post 20 Oct 2006, 02:11
A certain basket contains 10 apples, 7 of which are red and 3 of which are green. If 3 different apples are to be selected at random from the basket, what is the probability that 2 of the apples selected will be red and 1 will be green?

a. 7/40
b. 7/20
c. 49/100
d. 21/40
e. 7/10

OA is D. Can someone explain how you arrive at D? Thx
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 [#permalink] New post 20 Oct 2006, 05:03
My method is to look at all the possibilities one by one:

Probability: RRG=(7/10)*(6/9)*(3/8)=7/40
Probability: GRR=(3/10)*(7/9)*(6/8)=7/40
Probability: RGR=(7/10)*(3/9)*(6/8)=7/40

Add those probabilities together

(7/40)+(7/40)+(7/40)=21/40

D
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 [#permalink] New post 20 Oct 2006, 10:37
(7C2 * 3C1)/(10C3) = 63/120 = 21/40..........D
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 [#permalink] New post 20 Oct 2006, 10:45
Total cases = 10C3
Cases when 2 are red and one is green = 7C2 * 3C1

Prob = 7C2 * 3C1/ 10C3
= 21/40
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 [#permalink] New post 20 Oct 2006, 10:50
as above,

3 scenarios:

RRG,RGR,GRR

RRG=(7/10*6/9*3/8) = 7/40

others are same probability, so 7*3/40 = 21/40
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) [#permalink] New post 20 Oct 2006, 20:30
londonluddite wrote:
as above,

3 scenarios:

RRG,RGR,GRR

RRG=(7/10*6/9*3/8) = 7/40

others are same probability, so 7*3/40 = 21/40


Perfectly explained!
)   [#permalink] 20 Oct 2006, 20:30
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