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:



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