Question

Let a sample space be partitioned into three mutually exclusive and exhaustive events, B₁, B₂, and, B₃. Complete the following probability table. (Round your answers to 2 decimal places.)

Prior Probabilities			Conditional Probabilities			Joint Probabilities			Posterior Probabilities
P(B1)	=	0.11	P(A | B1)	=	0.45	P(A ∩ B1)	=		P(B1 | A)	=
P(B2)	=		P(A | B2)	=	0.62	P(A ∩ B2)	=		P(B2 |A)	=
P(B3)	=	0.38	P(A | B3)	=	0.85	P(A ∩ B3)	=		P(B3 | A)	=
Total	=					P(A)	=		Total	=

Answer 1

Answer:

Given,

P(B2) = 1 - P(B1) - P(B3)

= 1 - 0.11 - 0.38

= 0.51

Total = 1

P(A $\cap$ B1) = P(B1)*P(A|B1) = 0.11*0.45 = 0.0495

P(A $\cap$ B2) = P(B2)*P(A|B2) = 0.51*0.62 = 0.3162

P(A $\cap$ B3) = P(B3)*P(A|B3) = 0.38*0.85 = 0.323

P(A) = 0.0495 + 0.3162 + 0.323 = 0.6887

P(B1|A) = P(A $\cap$ B1)/P(A) = 0.0495/0.6887 = 0.0719

P(B2|A) = P(A $\cap$ B2)/P(A) = 0.3162/0.6887 = 0.4591

P(B3|A) = P(A $\cap$ B3)/P(A) = 0.323/0.6887 = 0.4690

Total = 1