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Question
A room has three sockets for lamps. From a collection 10 bulbs of which 6 are defective. At night a person selects 3 bulbs, at random and puts them in sockets. What is the probability that room is still dark
Solution
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 bulbs in 10C3 ways.
∴ n(S) = 10C3
Let event A: The room is dark
For event A to happen 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)`
= `(6 xx 5 xx 4)/(10 xx 9 xx 8)`
= `1/6`.
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