A number is a multiple of 8 if the last 3 digits of it, form a number which is a multiple of 8.
For example: 45288, 16840, 90024 are all multiples of 8, because 288, 840 and 024, the last 2 digits, are multiples of 8.
The total number of 8-digit numbers that we can form using {1,2,3,4,5,6} is
![6*6*6*6*6*6*6*6= 6^{8}](https://tex.z-dn.net/?f=6%2A6%2A6%2A6%2A6%2A6%2A6%2A6%3D%206%5E%7B8%7D%20)
, as we have 6 options for each digit.
The last 3 digits of multiple of 8, formed from the set {1,2,3,4,5,6} can be one of the following:
{112, 136, 144, 152, 216, 224,232, 256, 264
312, 336, 344, 352, 416, 424,432, 456, 464
512, 536, 544, 552, 616, 624,632, 656, 664} ,
which is 27 of them.
For each of these 3 digits fixed in the end, we can form
![6*6*6*6*6*1*1*1= 6^{5}](https://tex.z-dn.net/?f=6%2A6%2A6%2A6%2A6%2A1%2A1%2A1%3D%206%5E%7B5%7D%20)
numbers
so in total, there are
![27*6^{5}](https://tex.z-dn.net/?f=27%2A6%5E%7B5%7D)
numbers divisible by 8, whose digits are numbers in the set {1, 2, 3, 4, 5, 6}.
P(multiple of 8)=n(multiple of 8)/n(numbers that can be formed)=
![\frac{27*6^{5}}{6^{8}} = \frac{27}{6^{3}}= ( \frac{3}{6} )^{3} =( \frac{1}{2} )^{3}=0.125](https://tex.z-dn.net/?f=%5Cfrac%7B27%2A6%5E%7B5%7D%7D%7B6%5E%7B8%7D%7D%20%3D%20%20%5Cfrac%7B27%7D%7B6%5E%7B3%7D%7D%3D%20%28%20%5Cfrac%7B3%7D%7B6%7D%20%29%5E%7B3%7D%20%3D%28%20%5Cfrac%7B1%7D%7B2%7D%20%29%5E%7B3%7D%3D0.125)
Answer: 0.125