A box contains 10 bulbs , of which just three are defective . if 5 bulbs are drawn at random,find the probabiltiy that sample of 5 contains

(1) exactly one defective bulbs

(2) exactly two defective bulbs

(3) no defective bulbs

10 bulbs are in total, in which 3 bulbs are defective.
So drawing 5 bulbs at random is done in ¹⁰C₅ ways

1) Exactly one bulb is defective
It is done in ³C₁ * ⁷C₄
So the probability will be ( ³C₁ * ⁷C4)/¹⁰C₅

2) Exactly two bulbs are defective
It is done in ³C₂ * ⁷C₃
So the probability will be ( ³C₂ * ⁷C3)/¹⁰C₅

3).No defective bulbs
It is done in ³C₀ * ⁷C₅
So the probability will be ( ³C₀ * ⁷C5)/¹⁰C₅