Suppose Alice knows that Bob's pseudorandom key generator has a slight bias and that with probability $51\%$ it produces a key with more $1\text{'}$s than $0\text{'}$s$.$ Alice and Bob play the CI Game. Alice chooses messages m$0=0000$ and m$1=1111$ to Bob, who randomly chooses b in $\{0,1\}$ and encrypts mb$$ with a one$-$time pad using his pseudorandom key generator. He gives the ciphertext c$=$mb$$r $($where r is his pseudorandom key$)$ to Alice. Alice computes s$=$c$$m$0.$ If s has more $1\text{'}$s than $0\text{'}$s$,$ she guesses that b$=0.$ If not, she guesses that b$=1.$ For simplicity in the following, we assume that the message lengths are odd $($so there cannot be the same number of $1\text{'}$s and $0\text{'}$s$).($a$)$ Show that exactly one of m$0$r and m$1$r has more $1\text{'}$s than $0\text{'}$s$.($b$)$ Show that P$($s has more $1\text{'}$s $$b$=0)=\mathrm{.}51($c$)$ Show that P$($s has fewer $1\text{'}$s $$b$=1)=\mathrm{.}51($d$)$ Show that Alice has a probability $\mathrm{.}51$ of winning.