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

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by BTGmoderatorDC » Mon Oct 29, 2018 8:39 pm

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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

OA D

Source: GMAT Prep
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by Jay@ManhattanReview » Mon Oct 29, 2018 11:00 pm
BTGmoderatorDC wrote: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

OA D

Source: GMAT Prep
We have 7 red and 3 green apples

Probability that out of those 3, 2 will be red and 1 will be green = [(7C2)*(3C1)] / (10C3) = [(7.6)/(1.2)*3] / [(10.9.8)/(1.2.3)] = 21/40

The correct answer: D

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Tue Oct 30, 2018 5:34 am
BTGmoderatorDC wrote: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

OA D

Source: GMAT Prep
Jay has demonstrated a solution that involves counting techniques. Here's one that involves 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

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Wed Oct 31, 2018 5:30 pm
BTGmoderatorDC wrote: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.

Answer: D

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