'osterior probabilities are probabilities based on the outcome of the sample information. These can be computed by developing a table using the following process. 1. Enter the states of nature in the first column, the prior probabilities for the states of nature, P(IIS;), in the second column and the conditional probabilities in the third column. 2. In column 4 compute the joint probabilities by multiplying the prior probability values in column 2 by the corresponding conditional probabilities in column 3. 3. Sum the joint probabilities in column 4 to obtain the probability of the sample information I, P(I). 4. In column 5, divide each joint probability in column 4 by P(I) to obtain the posterior probabilities, P(s;II). The prior probabilities are given to be P(S₁) = 0.5, P(S₂) = 0.4, and P(S3) = 0.1. The conditional probabilities given each state of nature are P(I|s₁) = 0.1, P(I|s₂) = 0.05, and P(Ils 3) = 0.2. Use the given prior and conditional probabilities to compute the joint probabilities. States of Nature $1 $2 $3 Prior Probabilities P(s;) Conditional Probabilities P(I|s;) Joint Probabilities P(In s;) P(In S₁) = P(S₁)P(I|S₁) = 0.5(0.1) = P(S₁) = 0.5 P(S₂) = 0.4 P(S3) = 0.1 The sum of the Joint Probabilities column gives P(I) = P(I|S₂) = 0.1 P(I|S₂) = 0.05 P(I|S3) = 0.2
'osterior probabilities are probabilities based on the outcome of the sample information. These can be computed by developing a table using the following process. 1. Enter the states of nature in the first column, the prior probabilities for the states of nature, P(IIS;), in the second column and the conditional probabilities in the third column. 2. In column 4 compute the joint probabilities by multiplying the prior probability values in column 2 by the corresponding conditional probabilities in column 3. 3. Sum the joint probabilities in column 4 to obtain the probability of the sample information I, P(I). 4. In column 5, divide each joint probability in column 4 by P(I) to obtain the posterior probabilities, P(s;II). The prior probabilities are given to be P(S₁) = 0.5, P(S₂) = 0.4, and P(S3) = 0.1. The conditional probabilities given each state of nature are P(I|s₁) = 0.1, P(I|s₂) = 0.05, and P(Ils 3) = 0.2. Use the given prior and conditional probabilities to compute the joint probabilities. States of Nature $1 $2 $3 Prior Probabilities P(s;) Conditional Probabilities P(I|s;) Joint Probabilities P(In s;) P(In S₁) = P(S₁)P(I|S₁) = 0.5(0.1) = P(S₁) = 0.5 P(S₂) = 0.4 P(S3) = 0.1 The sum of the Joint Probabilities column gives P(I) = P(I|S₂) = 0.1 P(I|S₂) = 0.05 P(I|S3) = 0.2
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Transcribed Image Text:Posterior probabilities are conditional probabilities based on the outcome of the sample information. These can be computed by developing a table using the following process.
1. Enter the states of nature in the first column, the prior probabilities for the states of nature, P(I|s;), in the second column and the conditional probabilities in the
third column.
2. In column 4 compute the joint probabilities by multiplying the prior probability values in column 2 by the corresponding conditional probabilities in column 3.
3. Sum the joint probabilities in column 4 to obtain the probability of the sample information I, P(I).
4. In column 5, divide each joint probability in column 4 by P(I) to obtain the posterior probabilities, P(s¡|1).
The prior probabilities are given to be P(s₁) = 0.5, P(s₂) = 0.4, and P(S3) = 0.1. The conditional probabilities given each state of nature are P(I|s₁) = 0.1, P(I|s₂) = 0.05, and
P(I|S3) = 0.2.
Use the given prior and conditional probabilities to compute the joint probabilities.
States of Nature Prior Probabilities P(s;) Conditional Probabilities P(I|s;) Joint Probabilities P(In s;)
P(In S₁)
P(S₁)P(I|S₁)
0.5 (0.1)
S1
S2
$3
P(S₁) = = 0.5
P(S₂) =
P(S3) = = 0.1
= 0.4
The sum of the Joint Probabilities column gives P(I)
=
P(I|S₁) = = 0.1
P(I|S₂) = 0.05
P(I|S3) = 0.2
=
=
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