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# A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea

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Math Expert
Joined: 02 Sep 2009
Posts: 52231
A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea  [#permalink]

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22 Jul 2015, 01:14
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Difficulty:

5% (low)

Question Stats:

81% (01:13) correct 19% (00:52) wrong based on 165 sessions

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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.

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

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22 Jul 2015, 03:21
1
1
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  [#permalink]

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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  [#permalink]

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22 Jul 2015, 04:13
3
1
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.

Method-1

10 red jellybeans and 10 blue jellybeans

Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140
Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120

Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3

Probability = 120 / 1140 = 2/19

Method-2

Probability 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

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

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

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22 Jul 2015, 05:20
1
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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Math Expert
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A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea  [#permalink]

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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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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

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22 Jul 2015, 10:41
1
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  [#permalink]

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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  [#permalink]

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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 above

In "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  [#permalink]

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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.

Method-1

10 red jellybeans and 10 blue jellybeans

Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140
Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120

Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3

Probability = 120 / 1140 = 2/19

Method-2

Probability 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

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  [#permalink]

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23 Jul 2015, 09:01
1
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.

Method-1

10 red jellybeans and 10 blue jellybeans

Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140
Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120

Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3

Probability = 120 / 1140 = 2/19

Method-2

Probability 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

Hi GMATinsight,
if the prompt says with replacement, How can I use method 1 with combinations to solve the question?

Thanks

Method-1 of Combination is Preferably NOT used in case of "With Replacement"

in case of "with replacement"

Total outcomes = 10C1*10C1*10C1 = 10*10*10 = 1000

Total Outcomes = 20C1*20C1*20C1 = 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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Posts: 52231
Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea  [#permalink]

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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  [#permalink]

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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.

Method-1

10 red jellybeans and 10 blue jellybeans

Total Outcomes = No. of ways to choose 3 Jelly bean at random out of a total 20 jellybeans = 20C3 = 1140
Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10C3 = 120

Probability = Favourable Outcomes / Total Outcomes = 10C3 / 20C3

Probability = 120 / 1140 = 2/19

Method-2

Probability 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

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Re: A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybea &nbs [#permalink] 19 Dec 2018, 18:04
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