Question

{A1,A2,A3}{A1,A2,A3} is an event space for sample space SS with probabilities P[A1]=0.1,P[A2]=0.3,P[A3]=0.6.P[A1]=0.1,P[A2]=0.3,P[A3]=0.6. If EE is an event in SS with P[E|A1]=0.3,P[E|A2]=0.6,P[E|A1]=0.3,P[E|A2]=0.6, and P[E|A3]=0.3,P[E|A3]=0.3, compute the following.P(E)=
P(A1/E)=
P(A2/E)=
P(A3/E)=

Answer 1

Answer:

Step-by-step explanation:

Given that $P[A_1]=0.1,\\P[A_2]=0.3,\\P[A_3]=0.6$

Since total gives 1, we find that these three events are mutually exclusive and exhaustive.

E is one event such that

$P[E|A_1]=0.3\\P[E|A_2]=0.6,\\and P[E|A_3]=0.3$

$P(A1E) = P(E/A1)*P(A1) = 0.03\\P(A2E)                        = 0.18\\P(A3E)                        = 0.18\\P(E) = sum of these 3 = 0.39$

$P(A1/E) = P(A1E)/P(E) = \frac{0.03}{0.39} =0.0769\\P(A2/E) =\frac{0.18}{0.39} =0.4615\\P(A3/E)= 0.4615$