A box contains 14 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability distribution of the number of defective bulbs.

Question

Philoid · Accepted Answer

Let X be the random variable denoting the number of defective bulbs drawn in each draw. Since we are drawing a maximum of 3 bulbs at a time, So we can get at max 3 defective bulbs as total defective bulbs are 5.

∴ X can take values 0,1,2 and 3

P(X=0) = P(drawing no defective bulbs)

As we are finding probability for 0 defective bulbs ,so we will select all 3 bulbs

from 9 good bulbs.

n(s) = total possible ways =

∴ P(X=0) =

P(X=1) = P(drawing 1 defective bulbs and 2 good bulbs)

As we are finding probability for 1 defective bulbs ,so we will select 2 bulbs

from 9 good bulbs and 1 from 5 defective ones

∴ P(X=1) =

Similarly,

P(X=2) =

P(X=3) =

So, Probability distribution is given below: