Subject: General Aptitude · Chapter: Probability · Exam: · Marks: · Difficulty:
A box contains 20 electric bulbs, out of which 4 are defective. Two balls are chosen at random from this box. The probability that at least one of them is defective, is
(a)$\frac{4}{19}$
(b)$\frac{7}{19}$
(c)$\frac{12}{19}$
(d)$\frac{21}{95}$
(e)None of these
Answer
Answer (as printed): B
Explanation
$P$ (none is defective) $=\frac{{ }^{16} C_{2}}{{ }^{20} C_{2}}=\left(\frac{16 \times 15}{2 \times 1} \times \frac{2 \times 1}{20 \times 19}\right)=\frac{12}{19}$. $\mathrm{P}($ at least 1 is defective $)=\left(1-\frac{12}{19}\right)=\frac{7}{19}$.
Explanation as extracted from the printed page; notation may be imperfect.