ABC26GN5368 · Probability
Subject: General Aptitude · Chapter: Probability · Exam: 2009 · Marks: · Difficulty:
A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If 5 balls are picked up at random, what is the probability that at least one is blue?
(a)$\frac{137}{143}$
(b)$\frac{18}{455}$
(c)$\frac{9}{91}$
(d)$\frac{2}{5}$
(e)None of these
Answer
Explanation
Total number of balls $=(6+2+4+3)=15$. Let $E$ be the event of drawing 5 balls out of 9 non-blue balls $$={ }^{9} C_{5}={ }^{9} C_{(9-5)}={ }^{9} C_{4}=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}=126 .$$ And, $n(S)={ }^{15} C_{5}=\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}=3003$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{126}{3003}=\frac{6}{143} .$$ ∴ $\quad$ Required probability $=\left(1-\frac{6}{143}\right)=\frac{137}{143}$.
Explanation as extracted from the printed page; notation may be imperfect.
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