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) = 0.20, p(s2) = 0.57, and p(s3) = 0.23. with sample information i, p(i | s1) = 0.12, p(i | s2) = 0.04, and p(i | s3) = 0.20. 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.

Given : The three possible states of nature: S1, S2, and S3. The prior probabilities are P(S1) = 0.20  P(S2) = 0.57, and P(S3) = 0.23.

With sample information I, P(I | S1) = 0.12, P(I | S2) = 0.04, and P(I | S3) = 0.20.

Using the law of total probability , we have

Using Bayes theorem , we have

the answer is su.k.my? ? co.kc

6,250

step-by-step explanation:

to get the area, multiply 360 by 160.(57,600)

to find out how many bags of fertilizer it will take, first figure out how many square feet one bag can hold. divide 1000 by 8(125) to get the unit rate(how many square feet one pound of fertilizer can hold)and multiply it by 50(6250). if i haven't made a mistake, that is the answer.