Answer of question 2 of HW6: π(π2 = π |π1 = π , π2 = π’) = = = = π(π2 = π, π1 = π , π2 = π’) π(π1 = π , π2 = π’) π(π2 = π, π1 = π , π2 = π’) βπβ{π,π } π(π2 = π, π1 = π , π2 = π’) π(π2 = π’ | π2 = π, π1 = π )π(π2 = π, π1 = π ) βπβ{π,π ,π} π(π2 = π’ | π2 = π, π1 = π )π(π2 = π, π1 = π ) π(π2 = π’ | π2 = π, π1 = π )π(π2 = π|π1 = π )π(π1 = π ) βπβ{π,π ,π} π(π2 = π’ | π2 = π, π1 = π )π(π2 = π|π1 = π )π(π1 = π ) = π(π2 = π’ | π2 = π)π(π2 = π | π1 = π )π(π1 = π ) βπβ{π,π ,π} π(π2 = π’ | π2 = π)π(π2 = π | π1 = π )π(π1 = π ) = = π(π2 = π’ | π2 = π)π(π2 = π | π1 = π ) βπβ{π,π ,π} π(π2 = π’ | π2 = π)π(π2 = π | π1 = π ) 0.8 × 0.05 8 = β 0.2424 0.8 × 0.05 + 0.1 × 0.8 + 0.3 × 0.15 33 Note: Mentioning the π(π1 = π’) is irrelevant.
© Copyright 2025 Paperzz