(a) Construct a probability model of the situation. State the inspectors determina tions and the Security Chief's conclusion as probabilities Solution. Let the sample space S be a set of days and nights. Define the following three events D=A secret document disappears X=Agent X is at headquarters A= It is d In these terms Inspector am say Pr(DnX| A)=Pr (D A). Pr(X| A) Inspect Pm says: Pr(DnX| A)=Pr(DI A)- Pr(X|A) And the Security Chief concludes (D∩X)=Pr(D)·Pr(X) (b) Is the Security Chiefs reasoning correct? Justify your answer. Solution. The security chief is wrong. For example, suppose that S consists of a single day and a single night S=day, night) Assign night and day each probability 1/ 2. Now suppose that Agent X is around during the night and a document disappears only at nigl X=night A=(day) Furthe tent with the inspectors'determinations Pr(D∩X|4) Pr(D∩X∩A Pr(a) Pr(D∩A)Pr(X∩A) Pr(D A) Pr(X\A)=Pr(A) Pr(A) Pr(D∩X∩A (DOXIA Pr(D A).Pr(X(=Pr(DnA. Pr(XnA=1� � � � � � � � � � � � � � 4 Problem Set 10 (a) Construct a probability model of the situation. State the inspectors’ determina￾tions and the Security Chief’s conclusion as probabilities. Solution. Let the sample space S be a set of days and nights. Define the following three events: D = A secret document disappears X = Agent X is at headquarters A = It is daytime. In these terms, Inspector AM says: Pr (D ∩ X A| ) = Pr (D | A) · Pr (X A| ) Inspect PM says: Pr D ∩ X A | = Pr D | A · Pr X | A And the Security Chief concludes: Pr (D ∩ X) = Pr (D) · Pr (X) (b) Is the Security Chief’s reasoning correct? Justify your answer. Solution. The security chief is wrong. For example, suppose that S consists of a single day and a single night: S = {day, night} Assign night and day each probability 1/2. Now suppose that Agent X is around during the night and a document disappears only at night: D = {night} X = {night} A = {day} Furthermore, suppose Pr (day) = Pr (night) = 1/2. These suppositions are consis￾tent with the inspectors’ determinations: Pr (D ∩ X A) = Pr (D ∩ X ∩ A) | = 0 Pr (A) Pr (D | A) · Pr (X A) = Pr (D ∩ A) Pr (X ∩ A) | = 0 Pr (A) · Pr (A) Pr D ∩ X A = Pr D ∩ � X � ∩ A | = 1 Pr A � � � � Pr X ∩ A Pr D | A Pr X | A = Pr D � ∩ � A · � � = 1 Pr A · Pr A