Let XXX be a random variable such that P(X=k)=12kP(X=k) = \frac{1}{2^k}P(X=k)=2k1 for k=1,2,…k=1, 2, \dotsk=1,2,…. Calculate the hazard function h(k)=P(X=k∣X≥k)h(k) = P(X=k | X \ge k)h(k)=P(X=k∣X≥k).
1/21/21/2
1/k1/k1/k
1/2k1/2^k1/2k
k/2kk/2^kk/2k