A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If three balls are picked up at random, what is… · ABC26GN5361
Subject: General Aptitude · Chapter: Probability · Exam: Bank P.O. 2009
A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If three balls are picked up at random, what is the probability that none is yellow?
(a)$\frac{3}{455}$
(b)$\frac{1}{5}$
(c)$\frac{4}{5}$
(d)$\frac{44}{91}$
(e)None of these
Answer
Explanation
Total number of balls $=(6+2+4+3)=15$. Let $E$ be the event of drawing 3 non-yellow balls. Then, $n(E)={ }^{12} C_{3}=\frac{12 \times 11 \times 10}{3 \times 2 \times 1}=220$. Also, $n(S)={ }^{15} C_{3}=\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=455$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{220}{455}=\frac{44}{91} .$$
Explanation as extracted from the printed page; notation may be imperfect.
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