Subject: General Aptitude · Chapter: Probability · Exam: 2010 · Marks: · Difficulty:
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If three marbles are picked up at random, what is the probability that 2 are blue and 1 is yellow?
(a)$\frac{3}{91}$
(b)$\frac{1}{5}$
(c)$\frac{18}{455}$
(d)$\frac{7}{15}$
(e)None of these
Answer
Answer (as printed): C
Explanation
Total number of marbles $=(6+4+2+3)=15$. Let $E$ be the event of drawing 2 blue and 1 yellow marble. Then, $n(E)=\left({ }^{4} C_{2} \times{ }^{3} C_{1}\right)=\frac{4 \times 3}{2 \times 1} \times 3=18$. Also, $n(S)={ }^{15} C_{3}=\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=455$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{18}{455} .$$
Explanation as extracted from the printed page; notation may be imperfect.