3 electric bulbs are to be fitted in a room. The bulbs are chosen at random from 10 bulbs having 6 functioning bulbs. Find the probability that the room gets lighted.
Explanation:
There are 6 functioning bulbs and 4 defected bulbs.
P (room is lighted) = 1 – P (room is not lighted)
P (room is not lighted) = P (choosing all 3 defected bulbs)
Total Number of ways of choosing bulbs = 10C3
Number of ways of choosing defected bulbs = 4C3
∴ P (room is not lighted) = 4C3/10C3 = 4×3×210×9×8 = 1/30
∴ P (room is lighted) = 1 – 1/30 = 29/30
Hence, option (b).
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