There are total `10` bulbs out of which `3` are defective.
So, Probability when no bulb is defective,
`P(D = 0) = (C(7,3))/(C(10,3)) = (7*6*5)/(10*9*8) = 7/24`
Probability when one bulb is defective,
`P(D = 1) = (C(3,1)*C(7,2))/(C(10,3)) = (3*21)/(120) = 21/40`
Probability when two bulbs are defective,
`P(D = 2) = (C(3,2)*C(7,1))/(C(10,3)) = (3*7)/(120) = 7/40`
Probability when all three drawn bulbs are defective,
`P(D = 3) = (C(3,3))/(C(10,3)) = (1)/(120) = 1/120`
With these values, we can draw the probability distribution table.
Please refer to video to see the table.
