Five defective bo

X represents the number of defective bolts drawn.

∴ X can take values 0,1,2 or 3

∵ there are total 25 bolts (20 good + 5 defectives) bolts

n(S) = total possible ways of selecting 5 bolts =

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

P(X = 1) = P(selecting 1 defective bolt and 3 good bolts)

=

P(X = 2) = P(selecting 2 defective bolts and 2 good bolts)

=

P(X = 3) = P(selecting 3 defective bolts and 1 good bolt)

=

P(X = 4) = P(selecting 4 defective bolts and no good bolt)

=

Now we have p_i and x_i.

Following table represents the probability distribution of X :