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

  • 6

1) 5C1/10C5

2) 5C2/10C5

3) 10C10 - (5C1+5C2+5C3+5C4+5C5)

  • 2

I think you answer is wrong:

i think it should be 3C1 x 7C4 / 10C5

  • -2
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