Subject: General Aptitude · Chapter: Probability · Exam: 2010 · Marks: · Difficulty:
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If 4 marbles are picked up at random, what is the probability that at least one of them is blue?
(a)$\frac{4}{15}$
(b)$\frac{69}{91}$
(c)$\frac{11}{15}$
(d)$\frac{22}{91}$
(e)None of these
Answer
Answer (as printed): B
Explanation
Total number of marbles = (6 + 4 + 2 + 3) = 15. Let E be the event of drawing 4 marbles such that none is blue. Then, n(E) = number of ways of drawing 4 marbles out of 11 non-blue 11 11 × 10 × 9 × 8 = $\mathrm{C_{4}}$ = = 330. 4 × 3 × 2 ×1 15 15 × 14 × 13 × 12 And, n(S) = $\mathrm{C_{4}}$ = = 1365. 4×3×2×1 n(E) 330 22 ∴ P(E) = P( E ) = = = . n(S) 1365 91 22 69 ∴ Required probability = 1 − = . 91 91