TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW01.3

Independent events/disjoint events

For three events $A$, $B$ and $C$ we know that $A$ and $C$ are independent, $B$ and $C$ are independent, and $A$ and $B$ are disjoint. The probabilities $P(A\cup C)={\frac {2}{3}}$ , $P(B\cup C)={\frac {3}{4}}$ , $P(A\cup B\cup C)={\frac {11}{12}}$

are also given. Find $P(A)$, $P(B)$ and $P(C)$.

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

Lösungsvorschlag von Lessi

--Lessi 2024-02-07T13:04:11Z

Using Inclusion-Exclusion-Principle the given probabilities can be expanded: ${\begin{aligned}P(A\cup C)&=P(A)+P(C)-P(A\cap C)\\P(B\cup C)&=P(B)+P(C)-P(B\cap C)\\P(A\cup B\cup C)&=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C)\end{aligned}}$

Since A and B are disjoint $P(A\cap B)=P(A\cap B\cap C)=0$ we can expand $P(B\cap C)$ and $P(A\cap C)$ using the first two equations. After subtraction of probabilities we get: $P(C)=P(A\cup C)+P(B\cup C)-P(A\cup B\cup C)$

Since A and C are independent we can deduce $P(A)$ as follows: ${\begin{aligned}P(A\cup C)&=P(A)+P(C)-(P(A)*P(C))\\P(A\cup C)&=P(A)*(1-P(C))+P(C)\\P(A)&={\frac {P(A\cup C)-P(C)}{1-P(C)}}\end{aligned}}$

ac <- 2 / 3
bc <- 3 / 4
abc <- 11 / 12

c(ac, bc, abc)

# calculate P(C)
c <- ac + bc - abc

# calculate P(A)
a <- (ac - c) / (1 - c)

# calculate P(B)
b <- (bc - c) / (1 - c)

c(a, b, c)

# checking
a + c - (a * c)

b + c - (b * c)

a + b + c - (a * c) - (b * c)

${\begin{aligned}P(A)&={\frac {1}{3}}\\P(B)&={\frac {1}{5}}\\P(C)&={\frac {1}{5}}\end{aligned}}$