Having trouble figuring out how to remove duplicate counts in probability.

Question

curry · Answer

2.	There are 3 possible configurations to have 6 consecutive ones in a binary string of length 8.
1st Option – 6 ones from position 1 to 6.
2nd Option – 6 ones from position 2 to 7.
3rd Option – 6 ones from position 3 to 8.

However, need to exclude duplicate cases.

curry · Answer

@dan815

dan815 · Answer

okay

dan815 · Answer

when its all 1s it exists in all 3
when its 1s from 1 to 7 it exists in first 2
when its 1s fro 2 to 8 it exists in last 2

dan815 · Answer

2^8 - 6

curry · Answer

Is there a methodical way to think about this? or do i need to always jsut consider different cases?

dan815 · Answer

umm i dunno i just thouht about it like this

dan815 · Answer

so 
u can have all 1s there is 1 for this
7  of 1s-- there are 2 for this
6 of 1s -- there are 3 arrangements for htis

1+2+3=6

dan815 · Answer

oh wait there are more umm

curry · Answer

kk, that makes sense. Thank you!

dan815 · Answer

no theres more to it than just that let me think

dan815 · Answer

the 
8  of 1s
7 of 1s case is fine

dan815 · Answer

now for the 6 of 1s

dan815 · Answer

there are actually 5 cases

dan815 · Answer

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