Probability: 101

Nine marbles numbered from 1 to 9 are placed in a barrel and three are drawn out, without replacement. 


What is the probability that a3 digit number formed from the marbles in the order drawn, is divisible by 9? 

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Answer
5/42 i.e. 11.9 % 
As three marbles are drawn from nine marbles, without replacement, there are 9*8*7 = 504 possibilities. 
To be divisible by 9, the sum of the digits of any number must be divisible by 9. There are 10 such combinations with total 9 or 18. They are (1, 2, 6), (1, 3, 5), (2, 3, 4), (1, 8, 9), (2, 7, 9), (3, 6, 9), (3, 7, 8), (4, 5, 9), (4, 6, 8), (5, 6, 7) 
Each of these combinations can be arranged in 3! = 6 ways. Hence, total favourable cases are 6 * 10 = 60 
Thus, the probability, that a three digit number formed from the marbles in the order drawn is divisible by 9, is 60/504 = 5/42 = 11.9 % 
For a number to be divisible by 9, the order of the digits is not important. Hence the probability remains the same (5/42), if number is formed by rearranging the marbles drawn.