A basket contains 4 red, 5 blue and 3 green marbles. If three marbles are picked up at random what is the probability that at least one is blue ?

Answer: Option B

Total number of marbles = (4 + 5 + 3) = 12
Let E be the event of drawing 3 marbles such that none is blue.
Then, n (E) = number of ways of drawing 3 marbles out of 7 = $${}^7{C}_3$$   $$=\frac{7\times 6\times 5}{3\times 2\times 1}$$   = 35
And, $$n(S)={}^{12}{C}_3$$   $$=\frac{12\times 11\times 10}{3\times 2\times 1}$$   = 220
$$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{35}{220}=\frac{7}{44}$$
∴ Required probability
= 1 - P(E)
= $$\left(1-\frac{7}{44}\right)$$
= $$\frac{37}{44}$$