3 defective bulbs are mixed up with 7 good 3 bulbs are drawn at random. Find the probability distribution of defective bulbs.

Explanation

There are total `10` bulbs out of which `3` are defective.

So, Probability when no bulb is defective,

`P(D = 0) = (C(7,3))/(C(10,3)) = (7*6*5)/(10*9*8) = 7/24`

Probability when one bulb is defective,

`P(D = 1) = (C(3,1)*C(7,2))/(C(10,3)) = (3*21)/(120) = 21/40`

Probability when two bulbs are defective,

`P(D = 2) = (C(3,2)*C(7,1))/(C(10,3)) = (3*7)/(120) = 7/40`

Probability when all three drawn bulbs are defective,

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