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A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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08 Jul 2015, 03:10
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A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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08 Jul 2015, 03:38
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Bunuel wrote: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024 B. 1/24 C. 1/8 D. 105/512 E. 1/4
Kudos for a correct solution. Total balls in the Machine = 7+5+4 = 16METHOD1Probability of First ball to be Blue = 7/(7+5+4) = 7/16 Probability of Second ball to be Green = 5/(15) = 5/15 [15 balls remaining after one being taken out in Step 1] Probability of Third ball to be red = 4/(14) = 4/14 [14 balls remaining after Two being taken out in Step 1&2] But this can happen in any order so the orders in which it can happen = 3! i.e. Final Probability that machine dispenses one gumball of each color = (7/16)*(5/15)*(4/14)*3! = 1/4 Answer: Option E METHOD2The total ways in which three balls can be Picked out of a total of 16 balls (with orders) = 16*15*14 = 840 The ways in which three balls of different colors can be taken out (with orders) = 7*5*4*3! = 3360 Probability = Favourable Outcome / Total Outcomesi.e Probability = 840 / 3360 = 1/4 Answer: Option E METHOD3The total ways in which three balls can be Picked out of a total of 16 balls (without orders) = 16 C3 = 560 The ways in which three balls of different colors can be taken out (with orders) = 7 C1 * 5 C1 *4 C1 = 140 Probability = Favourable Outcome / Total Outcomesi.e Probability = 140 / 560 = 1/4 Answer: Option E
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A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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Updated on: 13 Jul 2015, 12:31
Bunuel wrote: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024 B. 1/24 C. 1/8 D. 105/512 E. 1/4
Kudos for a correct solution. Probability of selecting a red ball = P(R) = 4/16 Probability of selecting a green ball = P(G) = 5/15 Probability of selecting a blue ball = P(B) = 7/14 Total number of ways we can select the RBG combination =3! Thus the total probability is = (4/16)*(5/15)*(7/14)*3! = 1/4. Thus E is the correct answer.
Originally posted by ENGRTOMBA2018 on 08 Jul 2015, 04:17.
Last edited by ENGRTOMBA2018 on 13 Jul 2015, 12:31, edited 1 time in total.



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Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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08 Jul 2015, 07:21
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Total Events = 16C3 = 16*15*14/1*2*3 = 560 Probability = [7C1 * 5C1 * 4C1]/ 16C3 = 7 * 5 * 4/560 = 140/560 = 1/4. Ans (E).
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Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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08 Jul 2015, 21:49
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Bunuel wrote: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024 B. 1/24 C. 1/8 D. 105/512 E. 1/4
Kudos for a correct solution. Probability equation can be written as P(Differen color Balls = ( Number or ways Blue can be selected * Number or ways Green can be selected * Number or ways Red can be selected ) / total Number of ways 3 balls can be selected out of all. Total Number of Ways in which 3 balls can be selected out of 16 = 16C3 Total Number of ways in which 1 blue ball can be selected out of 7 blue balls = 7C1 = 7 Total Number of ways in which 1 Green ball can be selected out of 5 Green balls = 5C1 = 5 Total Number of ways in which 1 Red ball can be selected out of 4 Red balls = 4C1 = 4 Hence Required probability = (16*7*5) / 16C3 = 1/4 Option E should be correct answer.



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A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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13 Jul 2015, 12:14
Bunuel wrote: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024 B. 1/24 C. 1/8 D. 105/512 E. 1/4
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:Consider one specific case: blue first, then green, then red. By the dominoeffect rule, the probability of this case is: \(\frac{7 \ blue}{16 \ total}*\frac{5 \ green}{15 \ total}*\frac{4 red}{14 \ total}=\frac{1}{24}\). Now consider another case: green first, then red, then blue. The probability of this case is \(\frac{5 \ green}{16 \ total}*\frac{4 red}{15 \ total}*\frac{7 \ blue}{14 \ total}=\frac{1}{24}\). Notice that all we have done is swap around the numerators. We get the same final probability! This is no accident; the order in which the balls come out does not matter. Because the three desired gum balls can come out in any order, there are 3! = 6 different cases. All of these cases must have the same probability. Therefore, the overall probability is 6*1/24 = 1/4. In general, when you have a symmetrical problem with multiple equivalent cases, calculate the probability of one case (often by using the dominoeffect rule). Then multiply by the number of cases. Use combinatorics to calculate the number of cases, if necessary. Remember that when you apply a symmetry argument, the situation must truly be symmetrical. In the case above, if you swapped the order of "red" and "green" emerging from the gumball machine, nothing would change about the problem. As a result, we can use symmetry to simplify the computation.
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Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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14 Jul 2015, 05:00
Bunuel wrote: Bunuel wrote: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024 B. 1/24 C. 1/8 D. 105/512 E. 1/4
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:Consider one specific case: blue first, then green, then red. By the dominoeffect rule, the probability of this case is: \(\frac{7 \ blue}{16 \ total}*\frac{5 \ green}{15 \ total}*\frac{4 red}{ 15\ total}=\frac{1}{24}\). Now consider another case: green first, then red, then blue. The probability of this case is \(\frac{5 \ green}{16 \ total}*\frac{4 red}{15 \ total}*\frac{7 \ blue}{14 \ total}=\frac{1}{24}\). Notice that all we have done is swap around the numerators. We get the same final probability! This is no accident; the order in which the balls come out does not matter. Because the three desired gum balls can come out in any order, there are 3! = 6 different cases. All of these cases must have the same probability. Therefore, the overall probability is 6*1/24 = 1/3. In general, when you have a symmetrical problem with multiple equivalent cases, calculate the probability of one case (often by using the dominoeffect rule). Then multiply by the number of cases. Use combinatorics to calculate the number of cases, if necessary. Remember that when you apply a symmetry argument, the situation must truly be symmetrical. In the case above, if you swapped the order of "red" and "green" emerging from the gumball machine, nothing would change about the problem. As a result, we can use symmetry to simplify the computation. Bunuel: The Highlighted parts are representing mistakes The first Highlighted part 15 should have been 14 and The Second Highlighted part 6*1/24 = 1/3 should have been 6*1/24 = 1/ 4
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Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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14 Jul 2015, 05:05
GMATinsight wrote: Bunuel wrote: Bunuel wrote: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024 B. 1/24 C. 1/8 D. 105/512 E. 1/4
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:Consider one specific case: blue first, then green, then red. By the dominoeffect rule, the probability of this case is: \(\frac{7 \ blue}{16 \ total}*\frac{5 \ green}{15 \ total}*\frac{4 red}{ 15\ total}=\frac{1}{24}\). Now consider another case: green first, then red, then blue. The probability of this case is \(\frac{5 \ green}{16 \ total}*\frac{4 red}{15 \ total}*\frac{7 \ blue}{14 \ total}=\frac{1}{24}\). Notice that all we have done is swap around the numerators. We get the same final probability! This is no accident; the order in which the balls come out does not matter. Because the three desired gum balls can come out in any order, there are 3! = 6 different cases. All of these cases must have the same probability. Therefore, the overall probability is 6*1/24 = 1/3. In general, when you have a symmetrical problem with multiple equivalent cases, calculate the probability of one case (often by using the dominoeffect rule). Then multiply by the number of cases. Use combinatorics to calculate the number of cases, if necessary. Remember that when you apply a symmetry argument, the situation must truly be symmetrical. In the case above, if you swapped the order of "red" and "green" emerging from the gumball machine, nothing would change about the problem. As a result, we can use symmetry to simplify the computation. Bunuel: The Highlighted parts are representing mistakes The first Highlighted part 15 should have been 14 and The Second Highlighted part 6*1/24 = 1/3 should have been 6*1/24 = 1/ 4Edited the typos. Thank you.
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Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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12 Mar 2017, 09:28
GMATinsight wrote: Bunuel wrote: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumballs, which are identical except for their colors. If the machine dispenses three gumballs at random, what is the probability that it dispenses one gumball of each color?
A. 35/1024 B. 1/24 C. 1/8 D. 105/512 E. 1/4
Kudos for a correct solution. Total balls in the Machine = 7+5+4 = 16METHOD1Probability of First ball to be Blue = 7/(7+5+4) = 7/16 Probability of Second ball to be Green = 5/(15) = 5/15 [15 balls remaining after one being taken out in Step 1] Probability of Third ball to be red = 4/(14) = 4/14 [14 balls remaining after Two being taken out in Step 1&2] But this can happen in any order so the orders in which it can happen = 3! i.e. Final Probability that machine dispenses one gumball of each color = (7/16)*(5/15)*(4/14)*3! = 1/4 Answer: Option E METHOD2The total ways in which three balls can be Picked out of a total of 16 balls (with orders) = 16*15*14 = 840 The ways in which three balls of different colors can be taken out (with orders) = 7*5*4*3! = 3360 Probability = Favourable Outcome / Total Outcomesi.e Probability = 840 / 3360 = 1/4 Answer: Option E METHOD3The total ways in which three balls can be Picked out of a total of 16 balls (without orders) = 16 C3 = 560 The ways in which three balls of different colors can be taken out (with orders) = 7 C1 * 5 C1 *4 C1 = 140 Probability = Favourable Outcome / Total Outcomesi.e Probability = 140 / 560 = 1/4 Answer: Option E Can anyone please help me to understand why the orders of the balls considered important when calculating the number of desired outcomes? Thank you in advance. Regards, Nathan



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Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal [#permalink]
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Re: A miniature gumball machine contains 7 blue, 5 green, and 4 red gumbal
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