Buglary+System+Calculation+in+Bayesian+Network

Question: In the buglary system, who is more reliable wen a buglary happened? Mary or John? Proof it by calculation using bayesian Network Theory.

References of values: By using clausal inference in bayesian network: (from lecture notes)
 * P ( B )  ||  0.001  ||
 * P( ¬B)  ||  0.999  ||
 * P( E)  ||  0.002  ||
 * P( ¬E)  ||  0.998  ||
 * P(J|A) ||  0.90  ||
 * P( ¬J|A)  ||  0.10  ||
 * P(J| ¬A)  ||  0.05  ||
 * P( ¬J|¬A)  ||  0.95  ||
 * P(M|A) ||  0.70  ||
 * P( ¬M|A)  ||  0.30  ||
 * P(M| ¬A)  ||  0.01  ||
 * P( ¬ M| ¬ A)  ||  0.99  ||
 * P(A|B^E) ||  0.95  ||
 * P(A|B^ ¬E)  ||  0.94  ||
 * P(A| ¬ B^E)  ||  0.29  ||
 * P(A| ¬B^¬E)  ||  0.001  ||
 * P( ¬ A|B^E)  ||  0.05  ||
 * P( ¬ A| ¬ B^E)  ||  0.71  ||
 * P( ¬ A|B^ ¬ E)  ||  0.06  ||
 * P( ¬ A| ¬ B^ ¬ E)  ||  0.999  ||

John calls when there's buglary happened.

//P//(//J// | //B//) = //P//(//J// | //A// Λ //B//).//P//(//A// | //B//) + //P//(//J// | ¬//A// Λ //B//). //P//(¬//A// | //B//) = //P//(//J// | //A//).//P//(//A// | //B//) + //P//(//J// | ¬//A//).//P//(¬//A// | //B//) = //P//(//J// | //A//).//P//(//A// | //B//) + //P//(//J// | ¬//A//).(1 – //P//(¬//A// | //B//)) Alarm activated when there's buglary //P//(//A// | //B//) = //P//(//A// | //B// Λ //E//). P(//E// | //B//) + //P//(//A// | //B// Λ ¬//E//).//P//(¬//E// | //B//) = //P//(//A// | //B// Λ //E//).//P//(//E//) + //P//(//A// | //B// Λ ¬//E//).//P//(¬//E//) = 0.95 x 0.002 + 0.94 x 0.998 = 0.94002

//P//(//J// | //B//) = 0.90 x 0.94002 + 0.05 x 0.05998 = 0.849017

For Marry: Marry calls when there's buglary happened: //P//(M | //B//) = //P//(//M// | //A// Λ //B//).//P//(//A// | //B//) + //P//(//M// | ¬//A// Λ //B//). //P//(¬//A// | //B//) = //P//(//M// | //A//).//P//(//A// | //B//) + //P//(//M// | ¬//A//).//P//(¬//A// | //B//) = //P//(//M// | //A//).//P//(//A// | //B//) + //P//(//M// | ¬//A//).(1 – //P//(¬//A// | //B//))

//P//(//A// | //B//) = //P//(//A// | //B// Λ //E//). P(//E// | //B//) + //P//(//A// | //B// Λ ¬//E//).//P//(¬//E// | //B//) = //P//(//A// | //B// Λ //E//).//P//(//E//) + //P//(//A// | //B// Λ ¬//E//).//P//(¬//E//) = 0.95 x 0.002 + 0.94 x 0.998 = 0.94002

//P//(M | //B//) = 0.7 X 0.94002 + 0.01 X 0.05998 =0.6586138

Comparing the two values of probability, i get that the probability of John calls when there's buglary happened is higher ( 0.849017), compares to Marry calls when there's buglary happening (0.6586138). Therefore, John is more reliable when a buglary happened.

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