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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)

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Some Thoughts Probable solution | Comment 2 of 12 |

1a)  The number will only be divisible by 5 if the last digit is a 5, so the probability is 1/9
1b)  This is just a guess but I think one-seventh of the numbers are divisible by 7, so 1/7
1c)  The number will be divisible by 9 if its digits sum to 9 or 18.  There are 10 triples of digits that do so {126, 135, 234, 189, 279, 369, 378, 459, 468, 567}  and each has 3 arrangements so the probability is 60/504 = 5/42

2a) If any of the digits is a 5, so 3/9 = 1/3
2b) You get 6 arrangements for each trio, so 6/7
2c) is the same as 1c) 5/42


  Posted by Jer on 2006-04-25 13:04:13
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