If four marbles are picked at random, what is the probability that one is green, two are blue and one is red ?

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 4 marbles at random out of 15

=> $$P(S) = C^{15}_4 = \frac{15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4}$$

= $$1365$$

P(E) = Favorable outcomes

= Selecting 1 green, 2 blue and 1 red marble.

=> $$P(E) = C^2_1 \times C^6_2 \times C^3_1$$

= $$2 \times \frac{6 \times 5}{1 \times 2} \times 3$$

= $$90$$

$$\therefore$$ Required probability = $$\frac{P(E)}{P(S)}$$ 

= $$\frac{90}{1365} = \frac{6}{91}$$