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.
Well, since there's no divisibility rule for 7, here's what I did:
77 can only divide the number if there are an even number of 7s. Also, the number has to start with 77 because if it only started with 7, then how would it be divisible by 77? So, we could have the numbers:
There are 5 numbers. Each has a probability of (1/2)^5 of occuring, because the 7 in the hundred thousands place is already determined. So, I got 5/32, but the answer is actually 5/16? So what did I do wrong?
No, actually I was trying to find the cases that would work. You could have the number be 700000.
Yeah, but it's one of the possible cases, so it should be in the denom and counted.
I got 10 desirable outcomes out of a total of 32 possible options which is 10/32 = 5/16
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.
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