There are 10 solid colored balls in a box. . . . .

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

There are 10 solid colored balls in a box. . . . .

by Vincen » Fri Sep 29, 2017 7:14 pm

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There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A) 1/6
B) 7/30
C) 1/4
D) 3/10
E) 4/15

The OA is B.

Experts, could you please explain this using Combination and Probability approaches?

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Source: Beat The GMAT — Problem Solving | Fri Sep 29, 2017 7:14 pm

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Sat Sep 30, 2017 2:35 am

Vincen wrote:There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A) 1/6
B) 7/30
C) 1/4
D) 3/10
E) 4/15

From the 10 balls, 3 will be selected.
Thus, P(green is selected) = 3/10.
Of the 9 remaining balls, 7 will NOT be selected.
Thus, P(yellow is not selected) = 7/9.
To combine these probabilities, we multiply:
3/10 * 7/9 = 7/30.

The correct answer is B.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Sat Sep 30, 2017 2:49 am

Vincen wrote:There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A) 1/6
B) 7/30
C) 1/4
D) 3/10
E) 4/15

P(good outcome) = P(one way) * total possible ways.

Let G = green and N = not yellow.

P(one way):
One way to select the green marble but not the yellow marble is GNN.
P(G on the 1st pick) = 1/10. (Of the 10 marbles, 1 is green.)
P(N on the 2nd pick) = 8/9. (Of the 9 remaining marbles, 8 are not yellow.)
P(N on the 3rd pick) = 7/8. (Of the 8 remaining marbles, 7 are not yellow.)
Since we want all of these events to happen, we MULTIPLY:
1/10 * 8/9 * 7/8 = 7/90.

Total possible ways:
GNN is only ONE WAY to select the green marble but not the yellow marble.
Now we must account for ALL OF THE WAYS to select the green marble but not the yellow marble.
Any arrangement of the letters GNN represents one way to select the green marble but not the yellow marble.
Thus, to account for ALL OF THE WAYS to select the green marble but not the yellow marble, the result above must be multiplied by the number of ways to arrange the letters GNN.
Number of ways to arrange 3 elements = 3!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical N's:
3!/2! = 3.

Multiplying the results above, we get:
P(green but not yellow) = 3 * 7/90 = 7/30.

The correct answer is B.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Sat Sep 30, 2017 3:02 am

Vincen wrote:There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A) 1/6
B) 7/30
C) 1/4
D) 3/10
E) 4/15

P = (good combinations)/(all possible combinations)

All possible combinations:
From 10 marbles, the number of ways to select 3 = (10*9*8)/(3*2*1) = 120.

Good combinations:
To form a good combination, we must select a pair of non-yellow marbles to join the green marble.
Number of options for the 1st non-yellow marble = 8. (Of the 9 remaining marbles after the green marble has been selected, 8 are not yellow.)
Number of options for the 2nd non-yellow marble = 7. (Of the 8 remaining marbles, 7 are not yellow.)
To combine these options, we multiply:
8*7.
Since the ORDER of the two marbles does not matter -- BLUE_RED constitutes the same non-yellow pair as RED_BLUE -- we divide by the number of ways the two marbles can be ARRANGED (2!):
(8*7)/2! = 28.

Thus:
P = (good combinations)/(all possible combinations) = 28/120 = 7/30.

The correct answer is B.

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Mo2men: Legendary Member; Posts: 712; Joined: Fri Sep 25, 2015 4:39 am; Thanked: 14 times; Followed by:5 members

by Mo2men » Sat Sep 30, 2017 5:47 am

GMATGuruNY wrote:
Vincen wrote:There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A) 1/6
B) 7/30
C) 1/4
D) 3/10
E) 4/15
P = (good combinations)/(all possible combinations)

All possible combinations:
From 10 marbles, the number of ways to select 3 = (10*9*8)/(3*2*1) = 120.

Good combinations:
To form a good combination, we must select a pair of non-yellow marbles to join the green marble.
Number of options for the 1st non-yellow marble = 8. (Of the 9 remaining marbles after the green marble has been selected, 8 are not yellow.)
Number of options for the 2nd non-yellow marble = 7. (Of the 8 remaining marbles, 7 are not yellow.)
To combine these options, we multiply:
8*7.
Since the ORDER of the two marbles does not matter -- BLUE_RED constitutes the same non-yellow pair as RED_BLUE -- we divide by the number of ways the two marbles can be ARRANGED (2!):
(8*7)/2! = 28.

Thus:
P = (good combinations)/(all possible combinations) = 28/120 = 7/30.

The correct answer is B.

Dear Mitch,

Where is probability to pick the green ball? It is not clear from your solution above.

Can you help pls to understand?

Quote

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Sat Sep 30, 2017 5:58 am

Mo2men wrote:
Good combinations:
To form a good combination, we must select a pair of non-yellow marbles to join the green marble.
Number of options for the 1st non-yellow marble = 8. (Of the 9 remaining marbles after the green marble has been selected, 8 are not yellow.)
Number of options for the 2nd non-yellow marble = 7. (Of the 8 remaining marbles, 7 are not yellow.)
To combine these options, we multiply:
8*7.
Since the ORDER of the two marbles does not matter -- BLUE_RED constitutes the same non-yellow pair as RED_BLUE -- we divide by the number of ways the two marbles can be ARRANGED (2!):
(8*7)/2! = 28.
Dear Mitch,

Where is probability to pick the green ball? It is not clear from your solution above.

Can you help pls to understand?

Note the portion highlighted in blue.
The 28 pairs counted above constitute the number of pairs THAT CAN BE COMBINED WITH THE GREEN MARBLE to form a combination of 3 that includes the green marble but not the yellow marble.
Thus, there are 28 possible combinations of 3 that include the green marble but not the yellow marble.

Quote

Mo2men: Legendary Member; Posts: 712; Joined: Fri Sep 25, 2015 4:39 am; Thanked: 14 times; Followed by:5 members

by Mo2men » Sat Sep 30, 2017 11:58 am

GMATGuruNY wrote:
Mo2men wrote:
Good combinations:
To form a good combination, we must select a pair of non-yellow marbles to join the green marble.
Number of options for the 1st non-yellow marble = 8. (Of the 9 remaining marbles after the green marble has been selected, 8 are not yellow.)
Number of options for the 2nd non-yellow marble = 7. (Of the 8 remaining marbles, 7 are not yellow.)
To combine these options, we multiply:
8*7.
Since the ORDER of the two marbles does not matter -- BLUE_RED constitutes the same non-yellow pair as RED_BLUE -- we divide by the number of ways the two marbles can be ARRANGED (2!):
(8*7)/2! = 28.
Dear Mitch,

Where is probability to pick the green ball? It is not clear from your solution above.

Can you help pls to understand?
Note the portion highlighted in blue.
The 28 pairs counted above constitute the number of pairs THAT CAN BE COMBINED WITH THE GREEN MARBLE to form a combination of 3 that includes the green marble but not the yellow marble.
Thus, there are 28 possible combinations of 3 that include the green marble but not the yellow marble.

Thanks Mitch
But it is not clear how the picking of green marble included in your calculation? Where does this calculation appear?

Thanks

Quote

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Sat Sep 30, 2017 2:59 pm

Mo2men wrote:
GMATGuruNY wrote:
Mo2men wrote:
Good combinations:
To form a good combination, we must select a pair of non-yellow marbles to join the green marble.
Number of options for the 1st non-yellow marble = 8. (Of the 9 remaining marbles after the green marble has been selected, 8 are not yellow.)
Number of options for the 2nd non-yellow marble = 7. (Of the 8 remaining marbles, 7 are not yellow.)
To combine these options, we multiply:
8*7.
Since the ORDER of the two marbles does not matter -- BLUE_RED constitutes the same non-yellow pair as RED_BLUE -- we divide by the number of ways the two marbles can be ARRANGED (2!):
(8*7)/2! = 28.
Dear Mitch,

Where is probability to pick the green ball? It is not clear from your solution above.

Can you help pls to understand?
Note the portion highlighted in blue.
The 28 pairs counted above constitute the number of pairs THAT CAN BE COMBINED WITH THE GREEN MARBLE to form a combination of 3 that includes the green marble but not the yellow marble.
Thus, there are 28 possible combinations of 3 that include the green marble but not the yellow marble.
Thanks Mitch
But it is not clear how the picking of green marble included in your calculation? Where does this calculation appear?

Thanks

To form a combination of 3 that includes the green marble but not the yellow marble, we need to choose a PAIR OF NON-YELLOW MARBLES to COMBINE with the green marble.

Let the 8 non-yellow marbles be A, B, C, D, E, F, H, I.
From these 8 marbles, the following 28 pairs can be formed:
AB, AC, AD, AE, AF, AH, AI
BC, BD, BE, BF, BH, BI
CD, CE, CF, CH, CI
DE, DF, DH, DI
EF, EH, EI
FH, FI
HI

Each of these 28 pairs can be combined with the green marble to form a combination of 3, as follows:
ABG, ACG, ADG, AEG, AFG, AHG, AIG
BCG, BDG, BEG, BFG, BHG, BIG
CDG, CEG, CFG, CHG, CIG
DEG, DFG, DHG, DIG
EFG, EHG, EIG
FHG, FIG
HIG.
Total options = 28.

As illustrated above:
To count all of the possible 3-marble combinations that include the green marble but not the yellow marble, we can ignore the green marble.
What we must count is the number of PAIRS that can be formed from the 8 non-yellow marbles.
The reason:
Each of these non-yellow pairs can be combined with the green marble to form a viable combination of 3.

Quote

Mo2men: Legendary Member; Posts: 712; Joined: Fri Sep 25, 2015 4:39 am; Thanked: 14 times; Followed by:5 members

by Mo2men » Sat Sep 30, 2017 9:09 pm

GMATGuruNY wrote:
To form a combination of 3 that includes the green marble but not the yellow marble, we need to choose a PAIR OF NON-YELLOW MARBLES to COMBINE with the green marble.

Let the 8 non-yellow marbles be A, B, C, D, E, F, H, I.
From these 8 marbles, the following 28 pairs can be formed:
AB, AC, AD, AE, AF, AH, AI
BC, BD, BE, BF, BH, BI
CD, CE, CF, CH, CI
DE, DF, DH, DI
EF, EH, EI
FH, FI
HI

Each of these 28 pairs can be combined with the green marble to form a combination of 3, as follows:
ABG, ACG, ADG, AEG, AFG, AHG, AIG
BCG, BDG, BEG, BFG, BHG, BIG
CDG, CEG, CFG, CHG, CIG
DEG, DFG, DHG, DIG
EFG, EHG, EIG
FHG, FIG
HIG.
Total options = 28.

As illustrated above:
To count all of the possible 3-marble combinations that include the green marble but not the yellow marble, we can ignore the green marble.
What we must count is the number of PAIRS that can be formed from the 8 non-yellow marbles.
The reason:
Each of these non-yellow pairs can be combined with the green marble to form a viable combination of 3.

Dear Mitch

Your way above as I think resemble the following :

Probability (1 Green & 2 NON Yellow) = (1C1 ) (2C8)/ (3C10) = 7/30.

1C1 = 1, So it dose not affect the solution you presented above. Is my reasoning correct??

Thanks in advance

Quote

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Sun Oct 01, 2017 3:06 am

Mo2men wrote:
GMATGuruNY wrote:
To form a combination of 3 that includes the green marble but not the yellow marble, we need to choose a PAIR OF NON-YELLOW MARBLES to COMBINE with the green marble.

Let the 8 non-yellow marbles be A, B, C, D, E, F, H, I.
From these 8 marbles, the following 28 pairs can be formed:
AB, AC, AD, AE, AF, AH, AI
BC, BD, BE, BF, BH, BI
CD, CE, CF, CH, CI
DE, DF, DH, DI
EF, EH, EI
FH, FI
HI

Each of these 28 pairs can be combined with the green marble to form a combination of 3, as follows:
ABG, ACG, ADG, AEG, AFG, AHG, AIG
BCG, BDG, BEG, BFG, BHG, BIG
CDG, CEG, CFG, CHG, CIG
DEG, DFG, DHG, DIG
EFG, EHG, EIG
FHG, FIG
HIG.
Total options = 28.

As illustrated above:
To count all of the possible 3-marble combinations that include the green marble but not the yellow marble, we can ignore the green marble.
What we must count is the number of PAIRS that can be formed from the 8 non-yellow marbles.
The reason:
Each of these non-yellow pairs can be combined with the green marble to form a viable combination of 3.
Dear Mitch

Your way above as I think resemble the following :

Probability (1 Green & 2 NON Yellow) = (1C1 ) (2C8)/ (3C10) = 7/30.

1C1 = 1, So it dose not affect the solution you presented above. Is my reasoning correct??

Thanks in advance

Correct!
1C1 = the number of ways to choose the green marble.
Since 1C1 = 1, this value can be ignored when we count combinations that include the green marble but not the yellow marble.

Quote

Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Thu Jun 28, 2018 5:11 pm

Vincen wrote:There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A) 1/6
B) 7/30
C) 1/4
D) 3/10
E) 4/15

The number of ways to select 3 balls that include the green ball but not the yellow ball is:

1C1 x 1C0 x 8C2 = 1 x 1 x (8 x 7)/2! = 28 ways

(Note: 1C1 is the number of ways to select 1 green ball from the 1 green ball, 1C0 is the number of ways to select 0 yellow ball from the 1 yellow ball, and 8C2 is the number of ways to select 2 balls from the 8 remaining balls.)

The number of ways to select 3 balls from 10 is:

10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2) = 120

Thus, the probability that the 3 balls chosen will include the Green ball but not the yellow ball is 28/120 = 7/30.

Alternate Solution:

Let's first determine the probability of selecting the one green ball on the first draw and any of the remaining balls (except the yellow ball) on the succeeding two draws: 1/10 x 8/9 x 7/8 = 56/720.

Now, we see that we could also pick the green ball on the second draw, with any ball (except the yellow one) on the first and third draws, with probability: 8/10 x 1/9 x â…ž = 56/720.

Similarly, we could instead pick the green ball on the third draw, with any ball (except the yellow one) on the first and second draws, with probability: 8/10 x 7/9 x 1/8 = 56/720.

Thus the total probability for the event where the green ball is drawn but the yellow one is not is:
56/720 + 56/720 + 56/720 = 168/720 = 21/90 = 7/30.

Answer: B

Jeffrey Miller
Head of GMAT Instruction
[email protected]