Sum

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

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#### 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 ^{10}C_{3} ways.

∴ n(S) = ^{10}C_{3}

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 ^{6}C_{3} ways.

∴ n(A) = ^{6}C_{3}

∴ P(A) = `("n"("A"))/("n"("S"))`

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

= `(6 xx 5 xx 4)/(10 xx 9 xx 8)`

= `1/6`.

Concept: Basic Terminologies

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