Problem 4 23 algorithmic suppose that you are given a...

Problem 4-23 (Algorithmic) Suppose that you are given a decision situation with three possible states of nature: S1, S2, and S3. The prior probabilities are P(S1) = 0.24, P(S2) = 0.60, and P(S3) = 0.16. With sample information I, P(I | S1) = 0.10, P(I | S2) = 0.07, and P(I | S3) = 0.18. Compute the revised or posterior probabilities: P(S1 | I), P(S2 | I), and P(S3 | I). If required, round your answers to four decimal places. State of Nature P (Sj|I) S1 fill in the blank 1 0.2531 S2 fill in the blank 2 0.4430 S3 fill in the blank 3 0.3038 Total Probability fill in the blank 4

Transcript

00:01 So for this problem, we are given p of s1 equals 0 .24, p of s2 equals 0 .6, and p of s3 equals 0 .16.

00:08 As well as the conditional probabilities, p of i given s1 equals 0 .1, p of i given s2, equals 0 .0 .07, p of i given s3, equal 0 .18.

00:19 And we're asked to then find p of s1 given i, s2 given i, and s3 given i.

00:24 So what we want to do here is ultimately we'd be applying bayes ' theorem, which is going to be that the probability of b given a is equal to the probability of b and a divided by the probability of a.

00:47 But we would also have that the probability of a given b is equal to probability of b and a divided by the probability.

00:58 Of b.

01:00 So, ultimately, what we're going to want to do here is figure out the probability of b and a, so to speak, with each one of the conditional probabilities, then we can use that for the inverse probabilities.

01:13 So, particularly, we'd have that for i given s1, that's a probability of 0 .1, that must be equal to probability of i and s1, divided by probability of s1, which is, 0 .24.

01:32 So probability of i and s1 is equal to 0 .1 times 0 .24.

01:40 So that's 0 .024.

01:42 And we should be able to see the pattern here.

01:45 Essentially, our i and each possible state would be found by taking the i given that state and multiplying it by the probability of that state.

01:58 So we'd have 0 .07 times 0 .6 for i and s2 so that's 0 .042 and then probability of i and s3 equal to let's see here 0 .18 times 0 .0288 0 .088 0 .088 then let's see here so now that we have the combined probabilities here the ands we would have that the probability of i occurring is going to be equal to the sum of all of those ians so we have that the probability of i is going to be 0 .024 plus 0 .04 plus 0 .082 plus 0 .088 so now that we have that we can find the probability of s1 given i and so on by using the probability of i and s1, dividing that by the probability of i...

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