Difference between revisions of "TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2019W/1.3"

From VoWi
Jump to navigation Jump to search
m
(2 intermediate revisions by the same user not shown)
Line 8: Line 8:
 
}}
 
}}
  
{{ungelöst}}
+
== Lösungsvorschlag ==
 +
 
 +
Beweis für (a) von [https://math.stackexchange.com/questions/1962698 StackExchange]:
 +
 
 +
<math>
 +
\begin{align}
 +
P(A^c \cap B^c)
 +
&= 1 - P(A \cup B) \\
 +
&= 1 - P(A) - P(B) + P(A \cap B) \\
 +
&= 1 - P(A) - P(B) + P(A)P(B) \\
 +
&= (1-P(A))(1-P(B)) \\
 +
&= P(A^c)P(B^c).
 +
\end{align}
 +
</math>
 +
 
 +
Siehe auch diese [https://math.dartmouth.edu/archive/m20f10/public_html/ProofSolutions3.pdf PDF von math.dartmouth.edu].
 +
 
 +
(b): TBD

Revision as of 07:38, 13 October 2019

Independence

Let A and B be independent events.

(a) Prove that A^c and B^c are also independent.
(b) If we additionally know that P(A|B) = 0.6 and P(B|A) = 0.3, compute the probabilities of the following events
(i) at most one of A or B
(ii) either A or B but not both.

Lösungsvorschlag

Beweis für (a) von StackExchange:


\begin{align}
P(A^c \cap B^c) 
&= 1 - P(A \cup B) \\
&= 1 - P(A) - P(B) + P(A \cap B) \\
&= 1 - P(A) - P(B) + P(A)P(B) \\
&= (1-P(A))(1-P(B)) \\
&= P(A^c)P(B^c).
\end{align}

Siehe auch diese PDF von math.dartmouth.edu.

(b): TBD