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A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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28 Oct 2016, 09:08
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A bag contains 3 blue disks, 3 green disks and 4 orange disks. If three disks are selected random from the bag, what is the probability that 1 of the disks will be green and the other 2 of the disks will be orange? A) 1/30 B) 1/20 C) 3/40 D) 1/10 E) 3/20 My solution: Total disks: 10 The prob. of selecting a green=3/10 Once a green disk is selected, we are left with 9 total disks Then the probability of selecting an orange disk = 4/9 Once one orange is selected, we are left with 8 total disks Then the probability of selecting the second orange disk = 3/8
Total Probability = 3/10 * 4/9 * 3/8 ==> 1/20
However thats not the correct answer. What am I missing?
Appreciate your help!
Thanks
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A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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28 Oct 2016, 10:47
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HarveyKlaus wrote: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If three disks are selected random from the bag, what is the probability that 1 of the disks will be green and the other 2 of the disks will be orange? A) 1/30 B) 1/20 C) 3/40 D) 1/10 E) 3/20 My solution: Total disks: 10 The prob. of selecting a green=3/10 Once a green disk is selected, we are left with 9 total disks Then the probability of selecting an orange disk = 4/9 Once one orange is selected, we are left with 8 total disks Then the probability of selecting the second orange disk = 3/8
Total Probability = 3/10 * 4/9 * 3/8 ==> 1/20
However thats not the correct answer. What am I missing?
Appreciate your help!
Thanks Dear HarveyKlaus, I'm happy to respond. One very tricky thing about probability is that, when a question asked about the probability of a certain result, we have to be careful not to assume that there is only one way to get that result. For example, you approach explicitly assumes that the green disk was selected first. What if it wasn't? You calculated the probability of getting to that result by a specific route, but this excludes other ways to get to that same result. One way to approach this would be a kind of tree approachultimately, the choices, in order, could be GOO or OGO or OOG, and all three of those would have to be calculated and added. P(GOO) = 1/20 (what you calculated) P(OGO) = (4/10)*(3/9)*(3/8) = 1/20 P(OOG) = (4/10)*(3/9)*(3/8) = 1/20 Add P(total) = 3/20 Another approach would be to apply counting techniques. See: GMAT Probability and Counting TechniquesThink of it this way. Suppose the discs are give letters, A  J. Discs A, B, & C are green. Discs DG are orange, and rest are blue. Question #1: how many different sets of three altogether could we pick? 10C3 = \(\frac{10!}{(7!)(3!)}\) = \(\frac{10*9*8}{3*2*1}\) = 10*3*4 = 120. That's the denominator. Question #2: how many different sets of three involve just one green and two oranges? Three ways to choose the single green member, and 4C2 = 6 ways to choose the two orange members. By the Fundamental Counting Principle, number of ways = 3*6 = 18. That's the numerator. Probability = 18/120 = 3/20Here are a couple more blogs you may find helpful: GMAT Data Sufficiency Practice Questions on ProbabilityGMAT Advanced Probability ProblemsDoes this make sense? Mike
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Re: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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28 Oct 2016, 15:13
Thank you Mike! Thats very helpful! Best, Harvey mikemcgarry wrote: HarveyKlaus wrote: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If three disks are selected random from the bag, what is the probability that 1 of the disks will be green and the other 2 of the disks will be orange? A) 1/30 B) 1/20 C) 3/40 D) 1/10 E) 3/20 My solution: Total disks: 10 The prob. of selecting a green=3/10 Once a green disk is selected, we are left with 9 total disks Then the probability of selecting an orange disk = 4/9 Once one orange is selected, we are left with 8 total disks Then the probability of selecting the second orange disk = 3/8
Total Probability = 3/10 * 4/9 * 3/8 ==> 1/20
However thats not the correct answer. What am I missing?
Appreciate your help!
Thanks Dear HarveyKlaus, I'm happy to respond. One very tricky thing about probability is that, when a question asked about the probability of a certain result, we have to be careful not to assume that there is only one way to get that result. For example, you approach explicitly assumes that the green disk was selected first. What if it wasn't? You calculated the probability of getting to that result by a specific route, but this excludes other ways to get to that same result. One way to approach this would be a kind of tree approachultimately, the choices, in order, could be GOO or OGO or OOG, and all three of those would have to be calculated and added. P(GOO) = 1/20 (what you calculated) P(OGO) = (4/10)*(3/9)*(3/8) = 1/20 P(OOG) = (4/10)*(3/9)*(3/8) = 1/20 Add P(total) = 3/20 Another approach would be to apply counting techniques. See: GMAT Probability and Counting TechniquesThink of it this way. Suppose the discs are give letters, A  J. Discs A, B, & C are green. Discs DG are orange, and rest are blue. Question #1: how many different sets of three altogether could we pick? 10C3 = \(\frac{10!}{(7!)(3!)}\) = \(\frac{10*9*8}{3*2*1}\) = 10*3*4 = 120. That's the denominator. Question #2: how many different sets of three involve just one green and two oranges? Three ways to choose the single green member, and 4C2 = 6 ways to choose the two orange members. By the Fundamental Counting Principle, number of ways = 3*6 = 18. That's the numerator. Probability = 18/120 = 3/20Here are a couple more blogs you may find helpful: GMAT Data Sufficiency Practice Questions on ProbabilityGMAT Advanced Probability ProblemsDoes this make sense? Mike



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Re: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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28 Oct 2016, 16:00
Probability of Picking 1 green (3/10)*probability of orange (4/9)*probability of another orange (3/8)=1/20. There are three ways this can happen goo ogo oog 1/20*3=3/20



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Re: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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18 Dec 2016, 05:52
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Probability of complex events = Individual Events * their arrangements. In this case  it is equal to  Probability of 1 green  3/10 Probability of 1 orange  4/9 (Since 1 green has already selected.) Probability of 2nd orange  3/8 Their arrangements = 3!/2!*1! = 3 Probability = 3/10*4/9*3/8*3 = 3/20
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Re: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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18 Dec 2016, 14:15
Could you explain the part with " Their arrangements = 3!/2!*1! = 3 " ?
Thx



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Re: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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19 Dec 2016, 09:49
GMATforSeaCliff wrote: Could you explain the part with " Their arrangements = 3!/2!*1! = 3 " ?
Thx Sure. I meant the arrangements of the selected events i.e how many ways these events can arrange themselves. Total we have 3 events so the arrangement is 3! and out of them 2 are same i.e. selecting orange disk hence 3!/2! Hope its clear.
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Re: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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19 Dec 2016, 09:53
Thanks a lot Makes sense, just thought it the wrong way around Sent from my InFocus M808 using GMAT Club Forum mobile app



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Re: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If thre [#permalink]
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03 Jan 2017, 17:16
HarveyKlaus wrote: A bag contains 3 blue disks, 3 green disks and 4 orange disks. If three disks are selected random from the bag, what is the probability that 1 of the disks will be green and the other 2 of the disks will be orange?
A) 1/30 B) 1/20 C) 3/40 D) 1/10 E) 3/20 We are given that in a bag there are 3 blue disks, 3 green disks, and 4 orange disks. We need to determine the probability that from 3 selected disks, 1 will be green and 2 will be orange. Let’s first determine the total number of ways to select 1 green disk from 3. 3C1 = 3 Next, let’s determine the number of ways to select 2 orange disks from 4. 4C2 = (4 x 3)/2! = (4 x 3)/(2 x 1) = 6 ways So, there are 3 x 6 = 18 ways to select 1 green and 2 orange disks. Now let’s determine the number of ways to select 3 disks from 10: 10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2) = 5 x 3 x 8 = 120 Thus, the probability of selecting 1 green disk and 2 orange disks is 18/120 = 3/20. Alternate Solution: Selecting 3 disks at once is equivalent to selecting them one at a time without replacement. Remember, we are determining the probability of selecting 1 green disk and 2 orange disks from 10 total disks. If we select the green disk first, since there are 3 green disks and 10 total disks, there is a 3/10 chance that the green disk will be selected. Next, since there are 4 orange disks and 9 disks left, there is a 4/9 chance an orange disk will be selected. Similarly, for the third disk chosen, there is a 3/8 chance another orange disk will be selected. However, there are 3 different ways to select the 1 green disk and 2 orange disks: G  O  O O  G  O O  O  G Each of these 3 ways has the same probability of occurring. Thus, the total probability is: 3 x (3/10 x 4/9 x 3/8) = 3/20 Answer: E
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