Asked by qwerty
Let A,B,C be three events and let X=IA, Y=IB, Z=IC be the associated indicator random variables. We already know that X⋅Y is the indicator random variable of the event A∩B. In the same spirit, give an algebraic expression, involving X, Y, Z for the indicator random variable of the following events.
The event Ac∩Cc
Exactly one of the events A, B, C occurred.
The event Ac∩Cc
Exactly one of the events A, B, C occurred.
Answers
Answered by
RVE
1. (1-X)*(1-Z)
be aware your keyboard is set to US language not US international or other language
be aware your keyboard is set to US language not US international or other language
Answered by
Milorad
2) X*(1-X*Y)*(1-X*Z) + Y*(1-X*Y)*(1-Y*Z) + Z*(1-X*Z)*(1-Y*Z)
Answered by
123
2) Exactly one of the events A, B , C occurs:
[A ∩ (Bc ∩ Cc)] ∪ [B ∩ (Cc ∩ Ac)] ∪ [C ∩ (Ac ∩ Bc)]
Using 1) Indicator for the first formula: X*(1-Y)*(1-Z)
Now we have to use De Morgan Law and finally we have:
1-(1-X*(1-Y)*(1-Z))*(1-(1-X)*Y*(1-Z))*(1-(1-X)*(1-Y)*Z)
[A ∩ (Bc ∩ Cc)] ∪ [B ∩ (Cc ∩ Ac)] ∪ [C ∩ (Ac ∩ Bc)]
Using 1) Indicator for the first formula: X*(1-Y)*(1-Z)
Now we have to use De Morgan Law and finally we have:
1-(1-X*(1-Y)*(1-Z))*(1-(1-X)*Y*(1-Z))*(1-(1-X)*(1-Y)*Z)
Answered by
RVE
2. official answer:
1-(1-(X)*(1-Y)*(1-Z))*(1-(1-X)*(Y)*(1-Z))*(1-(1-X)*(1-Y)*(Z))
1-(1-(X)*(1-Y)*(1-Z))*(1-(1-X)*(Y)*(1-Z))*(1-(1-X)*(1-Y)*(Z))
Answered by
P
and if at most two of the events A, B, C occurred?
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