3. the first product has a probability of 70/100 to be selected from the acceptable ones. The second 69/99 (one is removed from the good ones) and the third one 68/98
4. so P(pick 3 consecutive acceptable products)==0.339
5. If familiar to the Combination formula C(n, r)=:
P(picking 3 acceptable out of 100)=n(picking 3 acceptable)/n(picking 3)=C(70, 3)/C(100/3)= = 0.339