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

true statements:

a) x + y = y + x   ⇒ commutative property of addition

b) (x * y) * z = x * (y * z) ⇒ associative property of multiplication

d) (x + y) + z = x + (y + z) ⇒ associative property of addition

false statements:

c) x - y = y - x   should be: x - y = - y + x

e) (x - y) - z = x - (y - z) should be: (x - y) - z = x - (y + z)

step-by-step explanation:

b. 24 mm

step-by-step explanation:

use the pythagorean theorem.

a^2 + b^2 = c^2

a^2 + 7^2 = 25^2

a^2 + 49 = 625

a^2 = 576

a = 24

answer: b. 24 mm