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

So drawing 5 bulbs at random is done in

^{10}C

_{5}ways

1) Exactly one bulb is defective

It is done in

^{3}C

_{1}*

^{7}C

_{4}

So the probability will be (

^{3}C

_{1}*

^{7}C4)/

^{10}C

_{5}

2) Exactly two bulbs are defective

It is done in

^{3}C

_{2}*

^{7}C

_{3}

So the probability will be (

^{3}C

_{2}*

^{7}C3)/

^{10}C

_{5}

3).No defective bulbs

It is done in

^{3}C

_{0}*

^{7}C

_{5}

So the probability will be (

^{3}C

_{0}*

^{7}C5)/

^{10}C

_{5}

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