Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 20 Jul 2019, 17:15

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 14 Sep 2010
Posts: 18
A certain basket contains 10 apples, 7 of which are red and  [#permalink]

### Show Tags

24 Jan 2011, 01:05
6
17
00:00

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
Joined: 02 Sep 2009
Posts: 56304
A certain basket contains 10 apples, 7 of which are red and  [#permalink]

### Show Tags

24 Jan 2011, 02:34
6
1
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.

_________________
Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 297

### Show Tags

05 Feb 2011, 07:38
6
1

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

### Show Tags

30 Oct 2018, 06:40
Top Contributor
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

RELATED VIDEO FROM OUR COURSE

_________________
Test confidently with gmatprepnow.com
Manager
Joined: 26 Feb 2017
Posts: 88
Location: India
GPA: 3.99
Re: A certain basket contains 10 apples, 7 of which are red and  [#permalink]

### Show Tags

30 Oct 2018, 06:48
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

Posted from my mobile device
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6967
Location: United States (CA)
Re: A certain basket contains 10 apples, 7 of which are red and  [#permalink]

### Show Tags

31 Oct 2018, 18:36
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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: A certain basket contains 10 apples, 7 of which are red and   [#permalink] 31 Oct 2018, 18:36
Display posts from previous: Sort by