Probability of an independent event according to past events A binary communication system is used to send one of two messages:
(i) message A is sent with probability 2/3, and consists of an infinite sequence of zeroes,
(ii) message B is sent with probability 1/3, and consists of an infinite sequence of ones.
The ith received bit is “correct" (i.e., the same as the transmitted bit) with probability 3/4, and is “incorrect" (i.e., a transmitted 0 is received as a 1, and vice versa), with probability 1/4. We assume that conditioned on any specific message sent, the received bits, denoted by Y1,Y2,… are independent.
Given that Y1,…,Y5 were all equal to 0, what is the probability that Y6 is also zero?
For me the answer was P(Y1 = 0) = P(Y6 = 0) because as I understood the received bits are independent, but the right answer is 0.749267.
Could you just help me to understand what I missed in the problem? :(
 A: Let us simplify notation, letting $X_{1-5}$ denote the event that the first five digits are all zero, and $X_6$ denote the event that the sixth digit is zero.
The question asks to find $Pr(X_6\mid X_{1-5})$, i.e. a conditional probability question asking us to find the probability that the sixth digit is zero given that the first five digits are all zero.
To do so, apply the definition of conditional probability:
$Pr(X_6\mid X_{1-5}) = \frac{Pr(X_{6}\cap X_{1-5})}{Pr(X_{1-5})}$
To continue, let us label a few more events.  $A$ is the event that the intended message to be sent is all zeroes, and $B$ will be the event that the intended message to be sent is all ones.  $Z_i$ will be the event that the $i^{th}$ digit is the intended digit.  Note that the problem statement tells us that each $Z_i$ is independent from $Z_j$ for each $i\neq j$.  I.e. the correctness of the digit is independent from digit to digit.  The value of the digit is most certainly not independent.
We note then that $X_6\cap X_{1-5} = \left(B\cap (Z_{1}^c\cap Z_2^c\cap\dots\cap Z_6^c)\right)\cup \left(A\cap (Z_1\cap Z_2\cap\dots\cap Z_6)\right)$
I.e. the message received has zeroes in the first six slots intentionally or unintentionally.
We continue then seeing $Pr(X_6\cap X_{1-5}) = \frac{1}{3}\cdot (\frac{1}{4})^6 + \frac{2}{3}\cdot (\frac{3}{4})^6$ and that $Pr(X_{1-5}) = \frac{1}{3}\cdot (\frac{1}{4})^5 + \frac{2}{3}\cdot (\frac{3}{4})^5$.
Completing the simplifications then, we arrive at $Pr(X_6\mid X_{1-5})=\frac{1459}{1948}\approx 0.74897$ 
calculations
A: The trick here is that we do not know if the message $A$ or $B$ was sent. If it's A, then $P(Y_6 = 0)= 0.75$ (correct bit). If it's B, then $P(Y_6 = 0)= 0.25$. 
But just calculating $P(A)*0,75 + P(B)*0,25 $ doesn't do the trick (it's roughly $0.417$), because this does not include our knowledge on the initial bits. We need to ask the question: "How likely is it that it's A given the first five bits are zero?"
Instead of $P(A)$ and $P(B)$, we need to compute the conditional probability of the message being A given that the first five bits were zeros and the conditional probability given that the first five bits were zeros, i.e. $P(A| Y_1 = Y_2 = Y_3 = Y_4 = Y_5=0)$ and $P(B| Y_1 = Y_2 = Y_3 = Y_4 = Y_5=0)$.
The problem gives us the probability for any bit being $0$ as $P(Y_i = 0 | A) = 0.75$ (if it's A, then $0$ is the correct bit) and $P(Y_i = 0|B) = 0.25$ (if it's B, then $0$ is the wrong bit).
With the help of Bayes' formula, can you complete the calculation?
The given answer $0.7492$ which is almost $0.75$ shows that it's pretty unlikely that the message is B (but it's still possible).
