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Three electric lamps are fitted in a room. 3 bulbs are chosen at random from 20 bulbs having 16 good
bulbs. The probability that the room is lighted is :

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$\frac{282}{285}$

$\frac{283}{285}$

$\frac{284}{285}$

$\frac{281}{285}$

answer is C.

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Detailed Solution

No of ways that 3 bulbs are chosen from 4 bad bulbs $=^{4} C_{3}$

Total possible cases $=^{20} C_{3}$

Required Probability $= 1 - \frac{^{4} C_{3}}{^{20} C_{3}} \Rightarrow 1 - \frac{1}{15 \times 19} = \frac{284}{285}$

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