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1. 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. $\frac{21}{95}$

Solution:

P (none is defective)
= n (E) =

\frac{^{16} C_{2}}{^{20} C_{2}} =

(\frac{16 \times 15}{2 \times 1} \times \frac{2 \times 1}{20 \times 19})

= \frac{12}{19}

P (at least 1 is defective)
=

(1 - \frac{12}{19})

= \frac{7}{19}

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