Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58445

A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 02:14
Question Stats:
82% (01:39) correct 18% (01:29) wrong based on 160 sessions
HideShow timer Statistics
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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 05:13
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
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION




Senior Manager
Joined: 28 Jun 2015
Posts: 283
Concentration: Finance
GPA: 3.5

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 04:21
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).
_________________
I used to think the brain was the most important organ. Then I thought, look what’s telling me that.



CEO
Joined: 20 Mar 2014
Posts: 2597
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
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. 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).



Director
Joined: 21 May 2013
Posts: 636

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 05:30
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



Manager
Joined: 22 Feb 2015
Posts: 56
Location: United States
Concentration: Finance, Operations
GMAT Date: 04012015
GPA: 3.98

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 06:20
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
_________________
Click +1 KUDOS , You can make me happy with just one click! Thanks



Math Expert
Joined: 02 Aug 2009
Posts: 7984

A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 07:58
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
_________________



Manager
Joined: 03 May 2014
Posts: 52
Concentration: Operations, Marketing
GMAT 1: 680 Q48 V34 GMAT 2: 700 Q49 V35
GPA: 3.6
WE: Engineering (Energy and Utilities)

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 11:41
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



Manager
Joined: 20 Jul 2011
Posts: 79

A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 12:22
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



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
WE: Education (Education)

A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
22 Jul 2015, 19:43
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)
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



SVP
Joined: 26 Mar 2013
Posts: 2344

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
23 Jul 2015, 09:42
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



CEO
Status: GMATINSIGHT Tutor
Joined: 08 Jul 2010
Posts: 2978
Location: India
GMAT: INSIGHT
WE: Education (Education)

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
23 Jul 2015, 10:01
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!
_________________
Prosper!!!GMATinsightBhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne Skype based classes and Classroom Coaching in South and West Delhihttp://www.GMATinsight.com/testimonials.htmlACCESS FREE GMAT TESTS HERE:22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION



Math Expert
Joined: 02 Sep 2009
Posts: 58445

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
26 Jul 2015, 12:14
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).
_________________



Manager
Joined: 21 Jul 2018
Posts: 186

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
19 Dec 2018, 19:04
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
_________________
______________________________ Press +1 Kudos if my post helped you a little and help me to ulcock the tests Wish you all success I'd appreciate learning about the grammatical errors in my posts
Please let me know if I'm wrong somewhere and help me to learn



GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4009
Location: Canada

Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
Show Tags
13 May 2019, 11:40
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. P(all 3 beans are blue) = P(1st bean is blue AND 2nd bean is blue AND 3rd bean is blue) = P(1st bean is blue) x P(2nd bean is blue) x P(3rd bean is blue) = 10/20 x 9/19 x 8/18 = 2/19 Answer: B Cheers, Brent
_________________
Test confidently with gmatprepnow.com




Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea
[#permalink]
13 May 2019, 11:40






