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A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 01:14
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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 03:21
Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. Sample space \(= 20C3 = \frac{20*19*18}{1*2*3} = 1140\). Favourable events \(= 10C3 = \frac{10*9*8}{1*2*3} = 120\). Probability \(= \frac{120}{1140} = \frac{2}{19}\). Ans (B).
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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 03:30
Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. required probability = 10C3 / 20C3 = 2/19. B is the correct answer. 10C3 = ways of selecting 3 blue ones out of 10 blue and 20C3 = ways of selecting 3 out of 20 (=10 red+10 blue).



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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 04:13
Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. Method110 red jellybeans and 10 blue jellybeans Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20 C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10 C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10 C3 / 20 C3 Probability = 120 / 1140 = 2/19 Answer: option B Method2Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 Answer: option B
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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 04:30
Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. Reqd. probability=(10/20)*(9/19)*(8/18) =2/19 Answer B



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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 05:20
We Have 10 Red Jellybeans and 10 Blue Jellybeans and our question is If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue? Total Outcomes = 20C3 = 1140 Fav. Outcomes = 10C3 = 120 Probability = Fav. Outcomes / Total Outcomes = 10C3/ 20C3 Probability = 120 / 1140 = 2/19 So B. 2/19 is the correct Answer Choice
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A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 06:58
Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. HI, the best and surest way to do is to calculate completely.. 10*9*8/(20*19*18)...either we calculate it as 2/19.. or if there is shortage of time for calculations, the denominator has 19, a prime number which is not there in numerator... so denominator should have 19 in it only B has it.. B
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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 10:41
Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. It is written with replacement. So for first blue 10/20, for second 9/19 and for 3rd 8/18..all together 10*9*8/20*19*18= 2/19



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A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 11:22
Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. Probability of selecting the first blue bean = 10/20 Probability of selecting the second blue bean = 9/19 Probability of selecting the third blue bean = 8/18 Hence Total Prob = (10/20) * (9/19) * (8/18) => 2/19 Option B



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A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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22 Jul 2015, 18:43
sahil7389 wrote: Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. It is written with replacement.So for first blue 10/20, for second 9/19 and for 3rd 8/18..all together 10*9*8/20*19*18= 2/19 Hi sahil7389, The Question says without Replacement. So I guess you have done a typo mistake as highlighted aboveIn "With Replacement" case calculation should be (10/20)*(10/20)*(10/20)
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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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23 Jul 2015, 08:42
GMATinsight wrote: Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. Method110 red jellybeans and 10 blue jellybeans Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20 C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10 C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10 C3 / 20 C3 Probability = 120 / 1140 = 2/19 Answer: option B Method2Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 Answer: option B Hi GMATinsight, if the prompt says with replacement, How can I use method 1 with combinations to solve the question? Thanks



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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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23 Jul 2015, 09:01
Mo2men wrote: GMATinsight wrote: Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. Method110 red jellybeans and 10 blue jellybeans Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20 C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10 C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10 C3 / 20 C3 Probability = 120 / 1140 = 2/19 Answer: option B Method2Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 Answer: option B Hi GMATinsight, if the prompt says with replacement, How can I use method 1 with combinations to solve the question? Thanks Method1 of Combination is Preferably NOT used in case of "With Replacement"in case of "with replacement" Total outcomes = 10 C1*10 C1*10 C1 = 10*10*10 = 1000 Total Outcomes = 20 C1*20 C1*20 C1 = 20*20*20 = 8000 Probability = 1000/8000 = 1/8 OR Probability = (10/20)*(10/20)*(10/20) = (1/2)^3 = 1/8 I hope it helps!
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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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26 Jul 2015, 11:14
Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. 800score Official Solution:To determine the probability that several events all take place, you must first determine the probability that each separate event occurs. In this case, to determine the probability of all 3 jellybeans being blue, you must first determine the separate probability of removing a blue jellybean on each trial. The first time a jellybean is removed, there are 10 blue jellybeans out of a total of 20 jellybeans in the bag. The probability of removing a blue one is 10/20. The second time a jellybean is removed, there are only 9 blue jellybeans out of a total of 19 (since we are only concerned with the case where the first was blue), so the probability of getting a blue one is 9/19. The third time, there are 8 blue jellybeans out of a total of 18 in the bag (by the same reasoning), so the probability is 8/18. To find the total probability, we must multiply the three probabilities together: 10/20 × 9/19 × 8/18. After reducing the fractions to 1/2 × 9/19 × 4/9, we get 2/19. In the last step, we reduce numerators and denominators first: the 9 in 9/19 cancels out with the 9 in 4/9, and the 2 in 1/2 cancels out when the 4 in 4/9 is reduced to 2. Now, the multiplication is 1 × 1/19 × 2, or 2/19. The correct answer is choice (B).
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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
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19 Dec 2018, 18:04
Hi GMATinsight, I solved this using method 2 specified in below mentioned solution, I am not aware in detail about 1st method by you, Could you please provide little detail about it so I can use that in exam if needed. Any one else can also assist me on this. GMATinsight wrote: Bunuel wrote: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?
A. 9/100 B. 2/19 C. 1/8 D. 3/20 E. 3/10
Kudos for a correct solution. Method110 red jellybeans and 10 blue jellybeans Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20 C3 = 1140 Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10 C3 = 120 Probability = Favourable Outcomes / Total Outcomes = 10 C3 / 20 C3 Probability = 120 / 1140 = 2/19 Answer: option B Method2Probability of First jelly bean to be Blue = 10/20 [Total 10 Blue out of total 20 jellybeans] Probability of Second jelly bean to be Blue = 9/19 [Total 9 Blue remaining out of total 19 jellybeans remaining] Probability of Third jelly bean to be Blue = 8/18 [Total 8 Blue remaining out of total 18 jellybeans remaining] Required Probability = (10/20)*(9/19)*(8/18) = 2/19 Answer: option B
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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea &nbs
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