[Algebraic Cryptography] How do you solve this? The second picture is definitions 5.20. Use the definition (5.15) of the probability of an event to prove thefollowing basic facts about probability theory.(a) Let E and F be disjoint events. ThenPr(EU F) = Pr(E) + Pr(F).(b) Let E and F be events that need not be disjoint. ThenPr(EUF) = Pr(E) + Pr(F) − Pr(E^F).(c) Let E be an event. Then Pr(Ec) = 1 − Pr(E).(d) Let E1, E2, E3 be events. Prove thatPr(E₁ U E2 U E3) = Pr(E₁) + Pr(E2) + Pr(E3) — Pr(E₁ ^ E₂)– Pr(E₁ E3) — Pr(E2 ~ E3) + Pr(E₁ E₂ E3).The formulas in (b) and (d) and their generalization to n events are knownas the inclusion-exclusion principle. Definition. A sample space (or set of outcomes) is a finite⁹ set . Eachoutcome we is assigned a probability Pr(w), where we require that theprobability functionPr : Ωsatisfy the following two properties:(a) 0≤ Pr(w) ≤ 1 for all w EN and (b) Σ Pr(ω) = 1. (5.14)WENRNotice that (5.14) (a) corresponds to our intuition that every outcomehas a probability between 0 (if it never occurs) and 1 (if it always occurs),while (5.14) (b) says that some outcome must occur, so contains all possibleoutcomes for the experiment.Definition. An event is any subset of 2. We assign a probability to an eventECN by setting(5.15)Pr(E) = Σ Pr(w).WEEIn particular, Pr(Ø) = 0 by convention, and Pr(N) = 1 from (5.14)(b).

As per the Bartleby guildlines we have to solve first three subparts and rest can be reposted....…

5.20. Use the definition (5.15) of the probability of an event to prove the following basic facts about probability theory. (a) Let E and F be disjoint events. Then Pr(EU F) = Pr(E) + Pr(F). (b) Let E and F be events that need not be disjoint. Then Pr(EU F) = Pr(E) + Pr(F) − Pr(E^ F). (c) Let E be an event. Then Pr(Ec) = 1 − Pr(E). (d) Let E1, E2, E3 be events. Prove that Pr(E₁ ¯ E2 ¯ E3) = Pr(E₁) + Pr(E2) + Pr(E3) − Pr(E₁ ^ E₂) – Pr(E₁ ^ E3) — Pr(E2 ^ E3) + Pr(E₁ ^ E2 ^ E3). The formulas in (b) and (d) and their generalization to n events are known as the inclusion-exclusion principle.