Let X be a binomial random variable with parameters we and p. Here is another way to prove that 1 1 2 n Pr[X is odd] = 5 (+33). LetX=X1+-~Xn where Pr[Xi=1]=pandPr[X,-=0]=1p. For each i, let Y} be a random variable with Pr[Kr = 1] = 2}) and Pr[Y; = O] = 1 230. Let Z,- be a random variable with Pr[Z;=1|Y.-=1]= Pr[Z;=0|Y.-=1]= NIH MID-I I and Pr[Zg=0|Y=0]=1. Intuitively: to determine Z;, we rst ip a coin with bias 2;). If the biased coin comes up tails, then Z,- is tails. If the biased coin comes up heads, then we ip a fair coin and use the fair coin's outcome for Z,-. Let Y = Y1 + + Yn and Z = Z1 + + Zn. Observe that Y is a binomial(n,2p) random variable and Z is a binomial(Y,%) random variable. Show that Z is a binomial(n,p) random variable by comparing the distributions of Z.- and X;. Therefore Pr[Z is odd] = Pr[X is odd]. The parity of Z is easily determined if we know whether Y is positive. If Y is 0, then Z must be 0. If Y > 0, then Z is a binomial(Y,%) random variable, so it must be even and odd with equal probability. Determine the probability that Z is odd by computing Pr[Z is oddA Y > D] + Pr[Z is odd A Y = 0]