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

Question

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.

A. 9/100
B. 2/19
C. 1/8
D. 3/20
E. 3/10

GMATinsight · Accepted Answer

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 = 20 C 3 = 1140
Favourable Outcomes = No. of ways to choose 3 Jelly bean such that they are all Blue out of 10 Blue = 10 C 3 = 120

Probability = Favourable Outcomes / Total Outcomes = 10 C 3 / 20 C 3

Probability = 120 / 1140 = 2/19

Answer: option B

Method-2

Probability of First jelly bean to be Blue = 10/20  
Probability of Second jelly bean to be Blue = 9/19  
Probability of Third jelly bean to be Blue = 8/18

Required Probability = (10/20)*(9/19)*(8/18) = 2/19

Answer: option B

BrentGMATPrepNow · Answer

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

TimeTraveller · Answer

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).

ENGRTOMBA2018 · Answer

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).

KS15 · Answer

Reqd. probability=(10/20)*(9/19)*(8/18)
 =2/19
Answer B

cutegirlsimran · Answer

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

chetan2u · Answer

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

sahil7389 · Answer

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

ashokk138 · Answer

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

GMATinsight · Answer

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)

Mo2men · Answer

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

Thanks

GMATinsight · Answer

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

in case of "with replacement"

Total outcomes = 10 C 1*10 C 1*10 C 1 = 10*10*10 = 1000

Total Outcomes = 20 C 1*20 C 1*20 C 1 = 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!

Bunuel · Answer

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).

Gmatprep550 · Answer

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.

ScottTargetTestPrep · Answer

The probability that 3 blue jellybeans will be removed without replacement is:

10/20 x 9/19 x 8/18

1/2 x 9/19 x 4/9 = 4/38 = 2/19

Answer: B

Vibhatu · Answer

Total 10 Blue out of total 20 jellybeans 
Probability of  First  jelly bean to be Blue = 10/20  
Probability of  Second  jelly bean to be Blue = 9/19  
Probability of  Third  jelly bean to be Blue = 8/18

Required Probability = (10/20)*(9/19)*(8/18) = 2/19

Answer: option B

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