Question
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 -
Answer: Option B
P (none is defective)
= n (E) = $$\frac{{{}^{16}\mathop C\nolimits_2 }}{{{}^{20}\mathop C\nolimits_2 }} = $$ $$\left( {\frac{{16 \times 15}}{{2 \times 1}} \times \frac{{2 \times 1}}{{20 \times 19}}} \right)$$ $$ = \frac{{12}}{{19}}$$
P (at least 1 is defective)
=$$\left( {1 - \frac{{12}}{{19}}} \right)$$ $$ = \frac{7}{{19}}$$
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P (none is defective)
= n (E) = $$\frac{{{}^{16}\mathop C\nolimits_2 }}{{{}^{20}\mathop C\nolimits_2 }} = $$ $$\left( {\frac{{16 \times 15}}{{2 \times 1}} \times \frac{{2 \times 1}}{{20 \times 19}}} \right)$$ $$ = \frac{{12}}{{19}}$$
P (at least 1 is defective)
=$$\left( {1 - \frac{{12}}{{19}}} \right)$$ $$ = \frac{7}{{19}}$$
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