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 3-digit results (9!/6!). The second is that for any unique 3-digit 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 3-digit 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 3-digit 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 2006-04-25 17:56:00