Question

A room has three sockets for lamps. From a collection of 10 light bulbs of which 6 are defective, a person selects 3 bulbs at random and puts them in socket. What is the probability that the room is lit?

Answer 1

Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 light bulbs in ¹⁰C₃ ways.
∴ n(S) = ¹⁰C₃
∴ Let A be the event that room is lit.
∴ A' is the event that the room is not lit.
For A' the bulbs should be selected from the 6 defective bulbs. This can be done in ⁶C₃ ways.
∴ n(A') = ⁶C₃
∴ P(A') = `("n"("A'"))/("n"("S"))=(""^6"C"_3)/ (""^10"C"_3)`
∴ P(Room is lit) = 1 − P(Room is not lit)
∴ P(A) = 1 – P(A')

= `1-(""^6"C"_3)/ (""^10"C"_3)`

= `1 - (6xx5xx4)/(10xx9xx8)`

= `1-1/6`

= `5/6`