Subject: General Aptitude · Chapter: Probability · Exam: 2010 · Marks: · Difficulty:
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If two marbles are picked up at random, what is the probability that either both are green or both are yellow?
(a)$\frac{5}{91}$
(b)$\frac{1}{35}$
(c)$\frac{1}{3}$
(d)$\frac{4}{105}$
(e)None of these
Answer
Answer (as printed): D
Explanation
Total number of marbles $=(6+4+2+3)=15$. Let $E$ be the event of drawing 2 marbles such that either both are green or both are yellow. Then, $n(E)=\left({ }^{2} C_{1}+{ }^{3} C_{2}\right)=\left(1+{ }^{3} C_{1}\right)=(1+3)=4$. And, $n(S)={ }^{15} C_{2}=\frac{15 \times 14}{2 \times 1}=105$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{4}{105} .$$
Explanation as extracted from the printed page; notation may be imperfect.