We have a lot of 6 items, 2 are defective →6−2=4 are non-defective.
P (drawing a defective item) = 26=13
P (drawing a non-defective item) = 1 - P (drawing a defective item) =1−13=23
Note:4 items can be drawn from a lot of 6 items in 6C4 ways.
Let X be the random variable of drawing 2 defective items from the lot.
P(X=0)= P (no defective item in the sample) = 4C46C4=115
P(X=1)= P (one defective item from the lot) = P(one defective, 3 non-defective items) =(2C14C36C4)=815
P(X=2)= P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =2C24C26C4=615
Therefore the required probability distribution is as follows:
X: 0 1 2
P(X):115 815 615