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

Question

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

Bunuel · Accepted Answer

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.

Answer: D.

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.

Answer: D.

dimitri92 · Answer

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

BrentGMATPrepNow · Answer

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

Answer: D

dollytaneja51 · Answer

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
Option D is the answer

Posted from my mobile device

ScottTargetTestPrep · Answer

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

Kinshook · Answer

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

Probability = 63/120 = 21/40

IMO D

Hovkial · Answer

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 =  /  =  21/40 
 
ANSWER: D

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.

Basshead · Answer

total outcomes = 10C3 = 120

favorable outcomes: 7C2 * 3C1 = 63

63/120 = 21/40

MathRevolution · Answer

Total Apples: 10

Red: 7 and Green: 3

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

2 Red  :  ^7{C_2} = 21  and 1 Green   =  ^3{C_1} =3

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

Answer D

Kinshook · Answer

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

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards