Case 3: 2 defectives { 1 , 2, { 1 , 3, = { 2, 3 , Thus, P(2 defectives)= Case 4: 3 defectives There are 7 such cases { 1,2,3 ,4}; { 1,2,3 ,5}; { 1,2,3 ,6}, { 1,2,3 ,7}, { 1,2,3 ,8}, { 1,2,3 ,9}, { 1,2,3 ,10} Thus, P(3 defectives) Thus probability distribution is X 1 2 3 P(x)=f(x) 1/6 1/2 3/10 1/30