Last visit was: 14 Sep 2024, 08:02 It is currently 14 Sep 2024, 08:02
Toolkit
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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# A certain basket contains 10 apples 7 of which are red and 3 are green

SORT BY:
Tags:
Show Tags
Hide Tags
Intern
Joined: 14 Sep 2010
Posts: 14
Own Kudos [?]: 294 [76]
Given Kudos: 0
Math Expert
Joined: 02 Sep 2009
Posts: 95503
Own Kudos [?]: 658500 [17]
Given Kudos: 87257
Manager
Joined: 15 Nov 2006
Affiliations: SPG
Posts: 232
Own Kudos [?]: 3238 [11]
Given Kudos: 34
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6803
Own Kudos [?]: 31311 [5]
Given Kudos: 799
A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
1
Kudos
4
Bookmarks
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

Originally posted by BrentGMATPrepNow on 30 Oct 2018, 06:40.
Last edited by BrentGMATPrepNow on 19 Sep 2021, 13:15, edited 1 time in total.
General Discussion
Manager
Joined: 26 Feb 2017
Posts: 70
Own Kudos [?]: 56 [1]
Given Kudos: 14
Location: India
GPA: 3.99
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
1
Kudos
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
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 19460
Own Kudos [?]: 23239 [0]
Given Kudos: 286
Location: United States (CA)
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
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.

GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5363
Own Kudos [?]: 4364 [0]
Given Kudos: 161
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
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

Total ways = 10C3 = 10*9*8/3*2 = 120
Favorable ways = 7C2*3C1 = 21*3 = 63

Probability = 63/120 = 21/40

IMO D
Director
Joined: 23 Apr 2019
Status:PhD trained. Education research, management.
Posts: 805
Own Kudos [?]: 2010 [0]
Given Kudos: 203
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
Probability of occurrence of an event E, P(E), is equal to the total number of favorable occurrences of that event, divided by, the total number of possible occurrences. That is,

P(E) = N(favorable)/N(total).

Calculate the numerator:

N(favorable) = Number of ways of choosing 2 red apples from 7 red apples, multiplied by, the number of ways of choosing 1 green apple from 3 green apples

N(favorable) = 7C2 * 3C1 ...............(1)

Calculate the denominator:

N(total) = Number of ways of choosing 3 apples from 10 apples, of which there are 7 red apples and 3 green apples = (10!)/(7!*3!) ........... (2)

Therefore, the required number of ways of choosing 2 red apples and 1 green apples = [7C2 * 3C1]/[(10!)/(7!*3!) ] = 21/40

Note: We are using the concept of combinations here since we are only interesting in choosing a set of elements from a larger set of elements. That is, we are not interested in the ordering of the chosen elements. There are two separate groups of identical elements in each group.
Director
Joined: 09 Jan 2020
Posts: 948
Own Kudos [?]: 239 [0]
Given Kudos: 432
Location: United States
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
total outcomes = 10C3 = 120

favorable outcomes: 7C2 * 3C1 = 63

63/120 = 21/40
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10131
Own Kudos [?]: 17278 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
1
Kudos
Total Apples: 10

Red: 7 and Green: 3

'3' apples are selected randomly, the total number of ways: $$^{10}{C_3} = 120$$

2 Red [From Red = 7]: $$^7{C_2} = 21$$ and 1 Green [From Green = 3] = $$^3{C_1} =3$$

=> Probability: $$\frac{(21 * 3) }{ 120} = \frac{21 }{40}$$

GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5363
Own Kudos [?]: 4364 [0]
Given Kudos: 161
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
Given: A certain basket contains 10 apples, 7 of which are red and 3 are green.
Asked: 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 ?

Total ways = 10C3 = 120
Favorable ways = 7C2*3C1 = 21*3 = 63
Probability = 63/120 = 21/40

IMO D
Non-Human User
Joined: 09 Sep 2013
Posts: 34832
Own Kudos [?]: 878 [0]
Given Kudos: 0
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: A certain basket contains 10 apples 7 of which are red and 3 are green [#permalink]
Moderator:
Math Expert
95503 posts