(s) A message is sent over a noisy channel. The message is a sequence $x_{1}, x_{2}, \ldots, x_{n}$ of $n$ bits $\left(x_{i} \in\{0,1\}\right)$. Since the channel is noisy, there is a chance that any bit might be corrupted, resulting in an error (a 0 becomes a 1 or vice versa). Assume that the error events are independent. Let $p$ be the probability that an individual bit has an error $(0<p<1 / 2)$. Let $y_{1}, y_{2}, \ldots, y_{n}$ be the received message (so $y_{i}=x_{i}$ if there is no error in that bit, but $y_{i}=1-x_{i}$ if there is an error there).
To help detect errors, the $n$th bit is reserved for a parity check: $x_{n}$ is defined to be 0 if $x_{1}+x_{2}+\cdots+x_{n-1}$ is even, and 1 if $x_{1}+x_{2}+\cdots+x_{n-1}$ is odd. When the message is received, the recipient checks whether $y_{n}$ has the same parity as $y_{1}+y_{2}+\cdots+y_{n-1}$. If the parity is wrong, the recipient knows that at least one error occurred; otherwise, the recipient assumes that there were no errors.
(a) For $n=5, p=0.1$, what is the probability that the received message has errors which go undetected?
(b) For general $n$ and $p$, write down an expression (as a sum) for the probability that the received message has errors which go undetected.
(c) Give a simplified expression, not involving a sum of a large number of terms, for the probability that the received message has errors which go undetected.
Hint for (c): Letting
$$
a=\sum_{k \text { even, } k \geq 0}\left(\begin{array}{l}
n \\
k
\end{array}\right) p^{k}(1-p)^{n-k} \text { and } b=\sum_{k \text { odd, } k \geq 1}\left(\begin{array}{l}
n \\
k
\end{array}\right) p^{k}(1-p)^{n-k}
$$
the binomial theorem makes it possible to find simple expressions for $a+b$ and $a-b$, which then makes it possible to obtain $a$ and $b$.