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 Three from Nine (Posted on 2006-04-25)
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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 Probable Solution | Comment 10 of 12 |

There are 9!/(9-3)! = 504 possible permutations of three-digit numbers of distinct digits from the set of digits 1 to 9.

(1a) 1/9; For any number of to be divisble by 5 formed from the set of digits 1 through 9 must end in a 5, and all numbers ending in 5 are divisible by 5. The probability of the last digit being a 5 is 1/9, therefore there is a 1/9 chance the number is divisible by 5.
(1b) 1/7
(1c) 5/42; To be divisible by 9, the sum of the digits of any number must be divisible by 9. For each of the 504/3! = 84 combinations, there are 10 combinations total 9 or 18, therefore there is a probability of 10*6/504 = 5/42 that the number is divisible by 9.
(2a) 1/3; There is a 1/3 chance that any one set of 3 marbles drawn will contain a 5. Therefore the probability that a number divisible by 5 can be formed is 1/3.
(2b) 9/14
(2c) 5/42; As in (1c), to be divisible by 9, the sum of the digits must be divisible by 9. The order of the digits is unimportant,. therefore the probability remains the same as in (1c).

Edited on April 25, 2006, 6:15 pm
 Posted by Dej Mar on 2006-04-25 16:17:09

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