A bag contains 10 red jellybeans

[Vincen]

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

The OA is B.

Should I use probability here? Could any expert help me?

[Brent@GMATPrepNow]

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

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

[EconomistGMATTutor]

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

The OA is B.

Should I use probability here? Could any expert help me?

Hi Vincen,
Let's take a look at your question.

This is a probability question and the jelly beans are removed one at a time from the bag and not replaced, so each time a jelly bean is removed from the bag the total number of jelly beans in the bag will be one less than the original number of jelly beans.

Let's first find the probability of removing the first blue jelly bean.
P(First jelly bean is blue) = 10/20 = 1/2

After removing the first blue jelly bean, now the total number of jelly beans in the bag are 19 out of which 9 are blue because one blue jelly bean is already removed from the bag.
P(Second jelly bean is blue) = 9/19

After removing the second blue jelly bean, now the total number of jelly beans in the bag are 18 out of which 8 are blue because two blue jelly beans are already removed from the bag.
P(Third jelly bean is blue) = 8/18 = 4/9

Now we will find the probability that all three jelly beans removed are blue.
P(All three jelly beans are blue) = P(First jelly bean is blue) x P(Second jelly bean is blue) x P(Third jelly bean is blue)
P(All three jelly beans are blue) = 1/2 x 9/19 x 4/9
P(All three jelly beans are blue) = (1 x 9 x 4)/(2 x 19 x 9)
P(All three jelly beans are blue) = 2/19

Therefore, Option B is correct.

I am available if you'd like any followup.

[Scott@TargetTestPrep]

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

The OA is B.

Should I use probability here? Could any expert help me?

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