Nine marbles numbered 1 to 9 are placed in a barrel and three are drawn out, without replacement. Determine :
1. The probability that the three digit number formed from the marbles in the order drawn is divisible by (a) Five (b) Seven (c) Nine.
2. The probability that a three digit number can be formed by rearranging the marbles drawn, that is divisible by (a) Five (b) Seven (c) Nine.
(In reply to
Partial Solution by tomarken)
tomarken's post : (c) No matter which two marbles you select first, there will only be one of the remaining seven that will make the number divisible by 9, so the answer is 1/7.
I thought this at first, too. But some combos don't work. If the first two were a 1 and 7, the third would have to be 1 but that isn't possible, so the answer is definitely less than 1/7.
Edited on April 25, 2006, 1:09 pm

Posted by Jer
on 20060425 13:08:44 