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

This topic has 6 expert replies and 3 member replies

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

Post Fri Sep 29, 2017 7:14 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post 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.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

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    Post 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.

    More practice:
    http://www.beatthegmat.com/select-exactly-2-women-t88786.html
    http://www.beatthegmat.com/probability-gmat-prep-t114250.html
    http://www.beatthegmat.com/a-single-particle-is-accelerated-through-a-magnetic-field-wh-t228342.html
    http://www.beatthegmat.com/at-a-blind-taste-competition-a-contestant-is-offered-3-cups-t220058.html
    http://www.beatthegmat.com/rain-check-t79099.html
    http://www.beatthegmat.com/probability-t227448.html

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

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    Post 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.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

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    Top Member

    Post 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?

    Post Sat Sep 30, 2017 5:58 am
    Mo2men wrote:
    Quote:
    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.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

    Thanked by: Mo2men
    Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.

    Top Member

    Post Sat Sep 30, 2017 11:58 am
    GMATGuruNY wrote:
    Mo2men wrote:
    Quote:
    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

    Post Sat Sep 30, 2017 2:59 pm
    Mo2men wrote:
    GMATGuruNY wrote:
    Mo2men wrote:
    Quote:
    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.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

    Thanked by: Mo2men
    Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.

    Top Member

    Post 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

    Post 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.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

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