Suppose that you are given a decision situation with three possible states of nature: S1, S2, and S3. The prior probabilities are P(S1) = 0.16, P(S2) = 0.57, and P(S3) = 0.27. With sample information I, P(I | S1) = 0.13, P(I | S2) = 0.07, and P(I | S3) = 0.19. 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.

26?

step-by-step explanation:

1010: 3030

is it i think

1. a^4+8a^3+24a^2+32a+16

2.y^8+8y^7+28y^6+56y^5+70y^4+56y^3+28y^2+8y+1

3.x^6-3x^4y^2+3x^2y^4-y^6

there is only one way to arrange all the eight relative to each other because they are not distinct. we have

now there are nine places to place an (and not more than one in one place).   because we have four symbols, we need to choose four of these nine places to insert an . order does not matter because the are not distinct.

thus there are 1 · c(9,4) ways, or, if we calculate it,

thus there are 126 ways to form such "words."