Subject: General Aptitude · Chapter: Probability · Exam: 2011 · Marks: · Difficulty:
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red ?
(a)$\frac{13}{35}$
(b)$\frac{24}{455}$
(c)$\frac{11}{15}$
(d)$\frac{1}{13}$
(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 1 green, 2 blue and 1 red marble. Then, $n(E)=\left({ }^{2} C_{1} \times{ }^{4} C_{2} \times{ }^{6} C_{1}\right)=2 \times \frac{4 \times 3}{2 \times 1} \times 6=72$. And, $n(S)={ }^{15} C_{4}=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=1365$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{72}{1365}=\frac{24}{455} .$$
Explanation as extracted from the printed page; notation may be imperfect.