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Bunuel
A bag contains 3 green marbles, three red marbles and 2 blue marbles. If 3 marbles are randomly drawn from the bag, what is the probability of selecting 2 red marbles and 1 green marble?

A. 1/336
B. 1/56
C. 3/56
D. 9/56
E. 9/28

Availability= 3G 3R 2B
Requirements: 2R 1G

Favourable cases = 3C2*3C1 = 3*3= 9
Total cases = 8C3 = 56

Probability= 9/56

Answer Option D
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Bunuel
A bag contains 3 green marbles, three red marbles and 2 blue marbles. If 3 marbles are randomly drawn from the bag, what is the probability of selecting 2 red marbles and 1 green marble?

A. 1/336
B. 1/56
C. 3/56
D. 9/56
E. 9/28

We are given that a bag contains 3 green marbles, 3 red marbles, and 2 blue marbles. We need to determine the probability of selecting 2 red marbles and 1 green marble.

The red marbles can be selected in 3C2 = 3 ways. The green marble can be selected in 3C1 = 3 ways.

The total number of ways to select the three marbles is 8C3 = 8!/[3!(8-3)!](8 x 7 x 6)/3! = 56.

Thus, the probability is (3 x 3)/56 = 9/56.

Answer: D
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Bunuel
A bag contains 3 green marbles, three red marbles and 2 blue marbles. If 3 marbles are randomly drawn from the bag, what is the probability of selecting 2 red marbles and 1 green marble?

A. 1/336
B. 1/56
C. 3/56
D. 9/56
E. 9/28

Bunuel,

I was trying to solve this as below, please let me know where I am going wrong.

Probability of selecting 2 red balls and 1 green ball,
= 3/8 x 2/7 x 3/6
= 3/56

Similarly probability of selecting 1 green and 2 red balls
= 3/8 x 3/7 x 2/6 = 3/56

Therefore, probability = 3/56 + 3/56 = 6/56


Thanks
Saurabh
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Sarjaria84
Bunuel
A bag contains 3 green marbles, three red marbles and 2 blue marbles. If 3 marbles are randomly drawn from the bag, what is the probability of selecting 2 red marbles and 1 green marble?

A. 1/336
B. 1/56
C. 3/56
D. 9/56
E. 9/28

Bunuel,

I was trying to solve this as below, please let me know where I am going wrong.

Probability of selecting 2 red balls and 1 green ball,
= 3/8 x 2/7 x 3/6
= 3/56

Similarly probability of selecting 1 green and 2 red balls
= 3/8 x 3/7 x 2/6 = 3/56

Therefore, probability = 3/56 + 3/56 = 6/56


Thanks
Saurabh

P(RRG) = 3/8*2/7*3/6*3!/2! = 9/56. We are multiplying by 3!/2! because RRG can occur in several way: RRG, RGR, GRR.

Answer: D
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Bunuel
Sarjaria84
Bunuel
A bag contains 3 green marbles, three red marbles and 2 blue marbles. If 3 marbles are randomly drawn from the bag, what is the probability of selecting 2 red marbles and 1 green marble?

A. 1/336
B. 1/56
C. 3/56
D. 9/56
E. 9/28

Bunuel,

I was trying to solve this as below, please let me know where I am going wrong.

Probability of selecting 2 red balls and 1 green ball,
= 3/8 x 2/7 x 3/6
= 3/56

Similarly probability of selecting 1 green and 2 red balls
= 3/8 x 3/7 x 2/6 = 3/56

Therefore, probability = 3/56 + 3/56 = 6/56


Thanks
Saurabh

P(RRG) = 3/8*2/7*3/6*3!/2! = 9/56. We are multiplying by 3!/2! because RRG can occur in several way: RRG, RGR, GRR.

Answer: D

So I missed solving for RGR selection, another 3/56 which will make total = 9/56

or 3/56*3!/2! = 9/56 (much faster)

Got it, Thanks Bunuel.
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