In a binary communication system all bit bytes are possible, and equally likely. Any bit is equallly likely to be a 1 or 0.
1- what is the probability of oberving a byte with exactly three 1's?
2-if a byte is observed with exactly three i's, what is the probability the first bit is a 1?
combination formula nCp
The Attempt at a Solution
1- I did:
P(exactly three 1's)= C(3,8)/ 2^8
Here I am confused:
I did P( three 1's and 1 is first)=C(2,7)/(2^8)
P(three 1's and 1 is first)=C(3,7)/(2^8)
I cannot see which one is right.
Can someone tell me what I did right and wrong?