Problem Solving

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

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

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.

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?

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.

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

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.