Subject: General Aptitude · Chapter: Probability · Exam: 2009 · Marks: · Difficulty:
A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If four balls are picked up at random, what is the probability that 2 are red and 2 are green ?
(a)$\frac{4}{15}$
(b)$\frac{5}{27}$
(c)$\frac{1}{3}$
(d)$\frac{2}{455}$
(e)None of these
Answer
Answer (as printed): D
Explanation
Total number of balls $=(6+2+4+3)=15$. Let $E$ be the event of drawing 4 balls such that 2 are red and 2 are green. Then, $n(E)=\left({ }^{2} C_{2} \times{ }^{4} C_{2}\right)=\left(1 \times \frac{4 \times 3}{2 \times 1}\right)=6$. And, $n(S)={ }^{15} C_{4}=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=1365$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{6}{1365}=\frac{2}{455} .$$
Explanation as extracted from the printed page; notation may be imperfect.