Probability of bit revelaed Eight bit string is received. First two bits are known to be 1. Any one out of the 8 bits is revealed. What is the probability of revealing a 1? I believe the probability is definitely skewed towards revealing a 1. P(revealing a 1) > P(revealing a 0). Is P(revealing a 1) = 5/8 and P(revealing a 0) = 3/8?
Thanks
 A: I have the same result, if p=0.5.
Probability choosing a "1"
a) Pobability choosing the first two bits is $\frac{2}{8}=\frac{1}{4}$ 
b) Probability choosing one of the bits 3-8 is  $\frac{6}{8}=\frac{3}{4}$
Now you have to calculate the probability, that among the 6 bits are k 1´s. This can be done by using the binomial distribution. This probability has to be multiplied by the probability of choosing a one, if k of 6 bits are 1´s. For every k the probability has to be calulated and summed up.
Alltogether it is $\frac{1}{4}+\frac{3}{4}\cdot \sum_{k=1}^6 \frac{k}{6} {6 \choose k} \cdot \left( \frac{1}{2} \right) ^k
 \cdot \left( \frac{1}{2} \right)^{6-k}$
A: Assuming that the revealed bit could be the first or second bit, known to be $1$, I would say the probability of seeing a $1$ is $\frac58$. Since I would expect $3$ of the unknown bits to be $1$ and three to be $0$, along with the two known bits of $1$.
[I am also assuming that the sent bits have a 50-50 chance each of being $1$ or $0$.]
