A box contains 4 defective and 6 non defective bulls. Find the probability that atleast 3 bulbs are defective when 4 bulbs are selected at random.

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We have `4` defective bulbs and `6` non-defective bulbs.

Now, we have to select `4` bulbs out of which at least `3` are defective.

So, we have to select either `3` defective bulbs or `4` defective bulbs.

When we have to select `3` defective bulbs, then probability `= (C(4,3)**C(6,1))/(C(10,4)`

When we have to select `3` defective bulbs, then probability `= (C(4,4)**C(6,0))/(C(10,4))`

So, required probability `P(E) = (C(4,3)**C(6,1)+C(4,4)**C(6,0))/(C(10,4))`

`=>P(E) = (4**6+1**1)/((10**9**8**7)/(4**3**2**1)) = 25/210 = 5/42`

So, required probability is `5/42`.