All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
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.)
Hints/Tips Partial Solution | Comment 1 of 12

Part One:

(a) For the three digit number to be divisible by 5, the last digit must be 5.  There is a 1/9 chance that the last digit will be 5, so the answer is 1/9.

(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.

Part Two:

(a) For this, any one of the three marbles must be a 5, so it is 1/9 + 1/9 + 1/9 = 1/3.  Alternately, this is 1 minus the probability that none of them is a 5, so 1 - 8/9*7/8*6/7 = 1 - 336/504 = 168/504 = 1/3.

(c) This is the same as in Part One.  If all the digits sum to nine, it doesn't matter what order they are in, this number will be divisible by 9, so the answer is still 1/7.

I haven't done part (b) of each, although I suspect that the answer is 1/7, at least for the first part.  I'll get to that one shortly...

  Posted by tomarken on 2006-04-25 12:30:48
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information