If four marbles are picked at random, what is the probability that at least one is yellow ?

Total number of marbles in the urn = 15

P(S) = Total possible outcomes

= Selecting 4 marbles at random out of 15

=> $$P(S) = ^{15}C_4 = \frac{15 \times 14 \times 13 \times 12}{1 \times 2 \times 3 \times 4}$$

= $$1365$$

Let no yellow marble is selected.

P(E) = Favorable outcomes

= Selecting 4 out of 11 marbles.

=> $$P(E) = ^{11}C_4$$

= $$\frac{11 \times 10 \times 9 \times 8}{1 \times 2 \times 3 \times 4}$$

= $$330$$

$$\therefore$$ Required probability = $$1 - \frac{P(E)}{P(S)}$$ 

= $$1 - \frac{330}{1365} = 1 - \frac{22}{91}$$

= $$\frac{91 - 22}{91} = \frac{69}{91}$$