2017 NS 19

Hint: A number is divisible by 77 when it is divisible by 7 and 11. There are a total of 2^6=64 total outcomes. 

Yep

I got 10 desirable outcomes out of a total of 32 possible options which is  10/32  =   5/16     

Method:

The first digit must be 7 then there are 5 more digits with 2 posibilities each so there are    2^5 = 32 possible numbers.

If a number is divisable by 77 then it must be divisable by 7 and 11.

Since all the digits are either 0 or 7 all the numbers will be divisable by 7 so I need to worry about if it is divisable by 11.

Now I looked up the internet fo find tricks for deciding if a number is divisable by 11.

This is what I found.

https://www.math.hmc.edu/funfacts/ffiles/10013.5.shtml

So if I talk about odd and even digit places then I have

7(even), odd,even, odd, even, odd

I have to have the same number of 7s in even numbers as I do in odd ones so that they will cancel each other out.

so I can have

6 sevens      one way to get this   777777

4 sevens      the first even seven is set, 2 choices for second even one, only one odd one gest left out so that is 3 choices,

                     so I have 1*2*3 = 6 ways to have 4 sevens.

2 sevens      The even one is set and there are 3 odd ones to chose from so that is  1*3 = 3

So altogether that makes     1+6+3 =10 numbers that are divisable by 77

So that is a probablility of   10/32   = 5/16