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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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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
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