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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 the room is lit
Solution
3 bulbs can be drawn from 10 bulbs in 10C3
= `(10xx9xx8)/(1xx2xx3)` = 120 ways
∴ n(S) = 120.
Let A = event that the room is dark.
∴ `bar"A"` ≡ event that room is lit
For A to happen, all 3 bulbs must be chosen from 6 defective bulbs. This can be done in 6C3 ways.
∴ n(A) = 6C3 = `(6xx5xx4)/(1xx2xx3)` = 20
∴ P (room is dark) = P(A) = `("n"("A"))/("n"("S"))=20/120=1/6`
The required probability = P (the room is lit)
= P(`bar"A"`)
= 1 – P(A)
= `1 - 1/6`
= `5/6`
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