Discussion

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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