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.
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