From a lot of 10

X represents the number of defective bulbs drawn.

∴ X can take values 0,1 or 2 (as maximum bulbs drawn are 2)

∵ there are total 10 bulbs (7 good+3 defectives)

n(S) = total possible ways of selecting 2 items from sample =

P(X = 0) = P(selecting no defective bulb) =

P(X = 1) = P(selecting 1 defective bulb and 1 good bulb)

=

P(X = 1) = P(selecting 0 defective bulb and 2 good bulb)

=

Now we have p_i and x_i.

Following table represents the probability distribution of X :