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.
The first premis is that there are 504 possible 3digit results (9!/6!). The second is that for any unique 3digit combination there are 6 permutations (3!) . I hope my stats knowledge serves me here because these values are used as the basis for some of the following answers.
1a) Only numbers ending in the digit 5 will be divisible by 5. Therefore the answer is equal to the probability that the last marble drawn will be the digit 5. (8/9 x 7/8 x 1/7 = 11.1%)
2a) If you are allowed to rearrange, then it is the probability that the digit 5 will be drawn first, second or third. This will be the inverse of the probability that it will not be drawn at all. (1  8/9 * 7/8 * 6/7 = 33.3%)
1c) and 2c) For divisibility by 9, being able to rearrange the result makes no difference, since the only criterion is that the sum of the digits drawn be a multiple of 9. There are 10 such 3digit combinations (shown below) which allows for 60 permutations. 60 / 504 = 11.9%
1+2+6 = 9
1+3+5 = 9
2+3+4 = 9
1+8+9 = 18
2+7+9 = 18
3+6+9 = 18
3+7+8 = 18
4+5+9 = 18
4+6+8 = 18
5+6+7 = 18
1a) Sorry, the only approach I had to this one was brute force. There are 72 3digit numbers that are divisible by 7 and do not contain the digit 0 or any duplicate digits. 72 / 504 = 14.3%
1b) Of the 72 possible numbers from 1a, there are only 54 unique digit combinations. The ability to rearrange the results allows for 324 permutations. 324 / 504 = 64.3%. That can't be right, can it?! Hmph, I guess it could be.
Edited on April 25, 2006, 6:04 pm
Edited on April 25, 2006, 6:20 pm

Posted by Kevin
on 20060425 17:56:00 