BSC channel probability, (binary symmetric channel) I have question regarding the binary symmetric channel (BSC), which assume each channel use is indepedent (i.e, if you send a '0', then you send '1', each time you send it is indepedent of others). Basically, the model of BSC is depicted as below. 

Assume we use repetition of length 3, i.e, if we want to send '0', we actually send '000'; if we want to send '1', we actually send '111'. 
Question 1). Given Bob receive '000', what is the probability that Alice send is '1'?
Question 2) What happen if we don't use repetition? I.e, we want to send '1', and we actually send a '1' instead of '111'. Given  Bob receive 000, what is the probaiblity Alice send '111'?
I got stuck on the second question. I think I have problem calculating $P(B=000)$ using the total probability theorem.

Attempt:
Question #1. 
$P(A=1|B=000)=P(B=000|A=111)*P(A=111)/ P(B=000)$
On the other hand, $P(B=000)=P(B=000|A=111)*P(A=111)+P(B=000|A=000)*P(A=000)$
using independence, we have $P(B=000)=P(B=0|A=1)^3*P(A=1)^3+P(B=0|A=0)^3*P(A=0)^3$
and all we need to do is to plug in $p$ into $P(B=0|A=1)$ and $P(B=1|A=0)$ into the equations.

Update 2:
For Case1: $P(B=000)=P(B=000|A=111)*P(A=111)+P(B=000|A=000)*P(A=000)$
However, for Case2:
$P(B=000)=P(B=000|A=000)*P(A=000)+P(B=000|A=001)*P(A=001)+P(B=000|A=010)*P(A=010).... +P(B=000|A=111)*P(A=111)$
 A: For the second problem, we assume that bits are sent with equal probability of being $0$ or $1$, and that they are independent. A very unreasonable assumption, unless Alice is sending gibberish. It is possible, I don't know Alice.
So given that $000$ was received, we find the probability $111$ was sent. Let $B$ be the event $000$ was received, and $A$ is the event $111$ was sent. We want $\Pr(A|B)$, which is $\frac{\Pr(A\cap B)}{\Pr(B)}$. 
We need to calculate a couple of probabilities. Easiest is $\Pr(A\cap B)$. The probability that $000$ was sent and $111$ received is $\frac{p^3}{8}$.
Now we find $\Pr(B)$. There are several cases. 
(i) $000$ was received and $111$ sent. We just did this.
(ii) $000$ was received and $000$ was sent. The probability is $\frac{(1-p)^3}{8}$.
(iii) One $0$ bit and two $1$'s were sent, and $000$ was received. The probability Alice sent a $0$ and two $1$'s is $\frac{3}{8}$. The probability the $0$ got through is $1-p$, and the probability the $1$'s were butchered is $p^2$, for a total of $\frac{(1-p)p^2}{8}$.
(iv) Two $0$ bits and a $1$ were sent, and $000$ received. The probability is $\frac{3(1-p)^2p}{8}$.
Add up the $4$ probabilities above to fin $\Pr(B)$.
