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Three from Nine (Posted on 2006-04-25) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Vernon Lewis    
Rating: 3.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Partial Solution | Comment 3 of 12 |
(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 2006-04-25 13:08:44

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