Problem 0.1 Let Xt be the number of people who enter a bank by time t gt; 0. Suppose Pr|Xt = k] = k! for k = 0, 1,2, ., and Pr[Xt = k, Xs = r] = _S'...

Please help me solve this 2 problems, thanks!

Problem 0.2 Recall the Geometric(p) distribution where X = number of flips of a coin until you get a head (H) with Pr(H) = p. The distribution is Pr(X = x) = (1 − p) x−1 p for x = 1, 2, . . . , with mean E(X) = ∑ ∞ x=1 x(1 − p) x−1 p = 1/p, which can be obtained by brute force. An easier way to find the mean is to condition on the first toss, say Y = 0 or 1 if the first toss is T or H. Show the mean is 1/p using E(X) = EE(X | Y).

Problem 0.1 Let Xt be the number of people who enter a bank by time t > 0. SupposePr|Xt = k] =k!for k = 0, 1,2, . . ., andPr[Xt = k, Xs = r] =_S' (t -s)k-re-tr!(k - r)!for t > s > 0, and k 2 r = 0,1, 2, . . .(a) Find Pr[X2 = k | X1 = 1] for k = 0, 1,2, . ...(b) Find E[X2 | X1 = 1].Useful information: Don't eat yellow snow, and e' = _fo t* /k!Problem 0.2 Recall the Geometric(p) distribution where X = number of flips of a coin until you get a head (H) withPr(H) = p. The distribution isPr(X = x) = (1 - p)*-1pfor x = 1,2,2,. . ., with mean E(X) = Er_1 x(1 - p)*-1p = 1/p, which can be obtained by brute force.An easier way to find the mean is to condition on the first toss, say Y = 0 or 1 if the first toss is T or H. Show themean is 1 / p using E(X) = EE(X | Y ).