Question

Show that for any three events \({\rm{A,B}}\), and \({\rm{C}}\) with \({\rm{P(C) > 0}}\),\({\rm{P(A}} \cup {\rm{B}}\mid {\rm{C) = P(A}}\mid {\rm{C) + P(B}}\mid {\rm{C) - P(A}} \cap {\rm{B}}\mid {\rm{C)}}\).

Short answer

The solution is

\(P(A \cup B\mid C) = P(A\mid C) + P(B\mid C) + P(A \cap B\mid C)\)

Step-by-step solution

Definition

The updated chance of an event occurring after additional information is taken into account is known as posterior probability.

Proofing that \({\rm{P}}\left( {{{\rm{B}}^{\rm{'}}}\mid {\rm{A}}} \right){\rm{ < P}}\left( {{{\rm{B}}^{\rm{'}}}} \right)\)

Remember the proposition:

Proposition: There are two events A and B for every two occurrences A and B.\({\rm{P(A}} \cup {\rm{B) = P(A) + P(B) - P(A}} \cap {\rm{B)}}\)

Given the occurrence of event B, the conditional probability of A is\({\rm{P(B) > 0}}\).

\({\rm{P(A}}\mid {\rm{B) = }}\frac{{{\rm{P(A}} \cap {\rm{B)}}}}{{{\rm{P(B)}}}}\)

Prove that \({\rm{P(A}} \cup {\rm{B}}\mid {\rm{C) = P(A}}\mid {\rm{C) + P(B}}\mid {\rm{C) + P(A}} \cap {\rm{B}}\mid {\rm{C)}}\)

For any two events \({\rm{A}}\)and.The following is true based on the definition and proposition:

\(\begin{aligned}P(A \cup B\mid C) &= P((A \cup B) \cap C) \\ &= \frac{{P((A \cap C) \cup (B \cap C))}}{{P(C)}}\frac{{P(A \cap C)}}{{P(C)}} \\ &= \frac{{P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)}}{{P(C)}} \\ &= \frac{{P(A \cap C)}}{{P(C)}} + \frac{{P(B \cap C)}}{{P(C)}} - \frac{{P((A \cap B) \cap C)}}{{P(C)}} \\ &= P(A\mid C) + P(B\mid C) + P(A \cap B\mid C) \\ \end{aligned} \)

(1): from the definition of conditional probability,

(2): this is true for any three events \(\backslash {\mathop{\rm Big}\nolimits} ((A \cup B) \cap C = (A \cap C) \cup (B \cap C)\backslash \) Big),

(3): We'll employ the above-mentioned proposition in this case.

(4): This is the inverse of the conditional probability definition.

Therefore, the solution is \(P(A \cup B\mid C) = P(A\mid C) + P(B\mid C) + P(A \cap B\mid C)\)