Assigned: Feb. 26
Due: Mar 5
Prob(P) = 0.8 Prob(C) = 0.19 Prob(H) = 0.01 Prob(K | P) = 0.1 Prob(K | C) = 0.2 Prob(K | H) = 1.0.A. Evaluate Prob(P | K), Prob(C | K), and Prob(H | K). Answer:
First, Prob(K) = Prob(K|P) Prob(P) + Prob(K|C) Prob(C) + Prob(K|H) Prob(H) = 0.1 * 0.8 + 0.2 * 0.19 + 1.0 * 0.01 = 0.08 + 0.038 + 0.01 = 0.128.
Prob(P | K) = (By Bayes' rule) Prob(P) Prob(K|P) / Prob(K) = 0.1 * 0.8 / 0.128 = 0.625
Prob(C | K) = (By Bayes' rule) Prob(C) Prob(K|C) / Prob(K) = 0.2 * 0.19 / 0.128 = 0.296
Prob(H | K) = (By Bayes' rule) Prob(H) Prob(K|H) / Prob(K) = 1.0 * 0.01 / 0.128 = 0.078
B. I check my pockets again, and again don't find the keys. Suppose that the two checks of my pockets are independent and identical. That is, let M be the event that I will miss my keys twice in checking the pocket twice. We suppose that
Prob(M | P) = (Prob(K | P))2 = 0.01 Prob(M | C) = (Prob(K | C))2 = 0.04 Prob(M | H) = (Prob(M | H))2 = 1.0.Evaluate Prob(P | M), Prob(C | M), and Prob(H | M).
First, Prob(M) = Prob(M|P) Prob(P) + Prob(M|C) Prob(C) + Prob(M|H) Prob(H) = 0.01 * 0.8 + 0.04 * 0.19 + 1.0 * 0.01 = 0.008 + 0.0076 + 0.01 = 0.0256
Prob(P | M) = (By Bayes' rule) Prob(P) Prob(M|P) / Prob(M) = 0.01 * 0.8 / 0.0256 = 0.311
Prob(C | M) = (By Bayes' rule) Prob(C) Prob(M|C) / Prob(M) = 0.04 * 0.19 / 0.0256 = 0.295
Prob(H | M) = (By Bayes' rule) Prob(H) Prob(M|H) / Prob(M) = 1.0 * 0.01 / 0.0256 = 0.394
C. Estimate how many times I have to check my pockets before I am 90% sure that I have left my keys at home. Answer: 4. (3 tries gives about 83% sure.)
A. Give an example to show that
P(Tag=noun | Tag=verb, Tag=conjunction)
is not zero. (All I'm asking for is one sentence with those three elements in a row.)
B. Propose a method that will allow you to combine probablistic information from the trigram model with syntactic constraints. Explain how this would fix this problem.