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Intern  Joined: 14 Sep 2010
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A certain basket contains 10 apples, 7 of which are red and  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 75% (01:54) correct 25% (02:02) wrong based on 734 sessions

### HideShow timer Statistics A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?

A. 7/40
B. 7/20
C. 49/100
D. 21/40
E. 7/10
Math Expert V
Joined: 02 Sep 2009
Posts: 56304
A certain basket contains 10 apples, 7 of which are red and  [#permalink]

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A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?

1) 7/40
2) 7/20
3) 49/100
4) 21/40
5) 7/10

Probability = # of favorable outcomes / total # of outcomes;

$$P=\frac{C^2_7*C^1_3}{C^3_{10}}=\frac{21}{40}$$, where $$C^2_7$$ is # of ways to choose 2 different green apples out of 7, $$C^1_3$$ is # of ways to choose 1 red apple out of 3, and $$C^3_{10}$$ is total # of ways to choose 3 different apples out of total 10 apples.

Or by probability approach: $$P(RRG)=\frac{3!}{2!}*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}=\frac{21}{40}$$, we are multiplying by 3!/2! as the case of GGR can occur in 3 ways: GGR, GRG, RGG - # of permutation of 3 letters out of which 2 are identical.

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probability of getting RRG

$$= \frac{7}{10}*\frac{6}{9}*\frac{3}{8} = \frac{7}{40}$$

there are 3 different ways to get 2 red and 1 green {RRG, RGR, GRR}.

so probability = $$\frac{7*3}{40} = \frac{21}{40}$$

Ans: D
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CEO  V
Joined: 12 Sep 2015
Posts: 3853
Re: A certain basket contains 10 apples, 7 of which are red and  [#permalink]

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A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?

A. 7/40
B. 7/20
C. 49/100
D. 21/40
E. 7/10

Let's use some probability rules.

Let's find the probability of selecting a red apple 1st, a red apple 2nd, and a green apple 3rd (aka RRG)
P(red apple 1st AND red apple 2nd AND green apple 3rd) = P(red apple 1st) x P(red apple 2nd) x P(green apple 3rd)
= 7/10 x 6/9 x 3/8
= 7/40

IMPORTANT: selecting a red apple 1st, a red apple 2nd, and a green apple 3rd (aka RRG) is JUST ONE WAY to get 2 red apples and 1 green apple. There's also RGR and GGR.

We already know that P(RRG) = 7/40, which also means P(RGR) = 7/40 and P(GRR) = 7/40
So, P(select 2 red apples and 1 green apple) = P(RRG OR RGR OR GRR)
= P(RRG) + P(RGR) + P(GRR)
= 7/40 + 7/40 + 7/40
= 21/40

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Re: A certain basket contains 10 apples, 7 of which are red and  [#permalink]

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Choosing 3 apples from a basket of 10 apples = 10C3
2 red apples and 1 green apple = 7C2 * 3C1
Probability = 7C2 * 3C1 / 10 C3
= 21 * 3 / 120
= 21/40

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Target Test Prep Representative D
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Re: A certain basket contains 10 apples, 7 of which are red and  [#permalink]

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A certain basket contains 10 apples, 7 of which are red and 3 are green. If 3 different apples are randomly selected, what is the probability that out of those 3 , 2 will be red and 1 will be green ?

A. 7/40
B. 7/20
C. 49/100
D. 21/40
E. 7/10

The number of ways to select 2 red apples is 7C2 = (7 x 6)/(2!) = 21.

The number of ways to select 1 green apple is 3C1 = 3.

So the total number of ways to select 2 red apples and 1 green apple is 21 x 3 = 63.

The total number of ways to select 3 apples from 10 is 10C3 = (10 x 9 x 8)/(3 x 2) = 5 x 3 x 8 = 120.

Thus, the total probability is 63/120 = 21/40.

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