Any of the 3! or 6 orders NDQ, NQD, DNQ, DQN, QND, QDN will be successful. The number of ways any one of these 6 orders can be selected is 3*2*5 or 30 ways. So there are 6*30 or 180 possible successful selections. The number of ways any three coins can be selected, since it is with replacement is 10*10*10 or 1000. Therefore the desired probability iswhich reduces to 9/500, or if expressed as a decimal is 0.018. Edwin