in this for quantity A, the probability will be 5/20 = 1/4. when it comes to quantity B, the probability is found out using P(A)*P(B) = (5/20)*(10/20) = 1/8

so the answer is A

I think both quantities are equal and answer is (C). This is my solution to this problem:

Let probability of selecting a green marble at random from the box is P(G) & probability of selecting a red marble at random from the box is P(R) & probability of selecting a marble at random from the box that is neither nor green is is P(neither G nor R)

P(G)*P(R) = 0, as both are mutually exclusive events.

Quantity A : P(G) = (5/20) =(1/4) ; B : P(neither G nor R) = 1 – (P(G)+P(R)) = 1 – ((5/20)+(10/20)) = 1/4

Answer is C

I also think C. There are 5 green marbles, 10 red marbles, and that means there must be 5 remaining that are neither red nor green. There is a 5/20 chance that the marble will be green. There is also a 5/20 chance it will be neither red nor green.

The solution is 'C' : Both quantities are equal

Explanation:
Total number of marbles = 20
Number of green marbles = 5
Number of red marbles = 10
Number of other colored marbles = 5
From a random draw on 20 marbles, probability of picking a green marble = 5/20 = 1/4 = P(G)
From a random draw on 20 marbles, probability of picking an other colored marble = 5/20 = 1/4 = P(OC)
We can see that P(G)=P(OC)=1/4

Hello Sandy,

I strongly doubt the validity of the answer to this being A. Kindly show how the answer being A is achieved.

Hi,

The answer is indeed C. Fixed it.
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