Subject: General Aptitude · Chapter: Probability · Exam: 2010 · Marks: · Difficulty:
A basket contains 4 red, 5 blue and 3 green marbles. If three marbles are picked up at random what is the probability that at least one is blue?
(a)$\frac{7}{12}$
(b)$\frac{37}{44}$
(c)$\frac{5}{12}$
(d)$\frac{7}{44}$
(e)None of these
Answer
Answer (as printed): B
Explanation
Total number of marbles $=(4+5+3)=12$. Let $E$ be the event of drawing 3 marbles such that none is blue. Then, $n(E)=$ number of ways of drawing 3 marbles out of $7={ }^{7} C_{3}=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35$. And, $n(S)={ }^{12} C_{3}=\frac{12 \times 11 \times 10}{3 \times 2 \times 1}=220$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{35}{220}=\frac{7}{44} .$$ Required probability $=1-P(E)=\left(1-\frac{7}{44}\right)=\frac{37}{44}$.
Explanation as extracted from the printed page; notation may be imperfect.