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Looking for a place to get what things equal

schinb65

New member
Jan 1, 2013
12
I am studying for an actuary exam. It has been years since I took a Stats/probability class. I am not really sure where I can find formulas that are the same.

I have this $Pr(a\cap h^c)$, so i believe that I can say $Pr(a\cap 1-h)$ is that true?

The solution has this:
$Pr((a\cap h^c)$ = $(a-h)$ = $(a-(a \cap h))$

So, if what I have is true, how can I go for $Pr(a\cap 1-h)$ to $(a-(a \cap h))$?

I was kind of hoping in the solution can be explained, or if any links where I would be able to see that the solution works as well as others so that I can review what I have forgotten over the years.

Thank you.
 

schinb65

New member
Jan 1, 2013
12
I am studying for an actuary exam. It has been years since I took a Stats/probability class. I am not really sure where I can find formulas that are the same.

I have this $Pr(a\cap h^c)$, so i believe that I can say $Pr(a\cap 1-h)$ is that true?

The solution has this:
$Pr((a\cap h^c)$ = $(a-h)$ = $(a-(a \cap h))$

So, if what I have is true, how can I go for $Pr(a\cap 1-h)$ to $(a-(a \cap h))$?

I was kind of hoping in the solution can be explained, or if any links where I would be able to see that the solution works as well as others so that I can review what I have forgotten over the years.

Thank you.
I think I got it, I can say this correct?
$Pr(A)= Pr(A\cap H) + Pr(a\cap H^c)$
Thus if I use algebra I can work it around to get $Pr(A)-Pr(A \cap H)=Pr((A\cap H^c)$
 

Amer

Active member
Mar 1, 2012
275
I am studying for an actuary exam. It has been years since I took a Stats/probability class. I am not really sure where I can find formulas that are the same.

I have this $Pr(a\cap h^c)$, so i believe that I can say $Pr(a\cap 1-h)$ is that true?

The solution has this:
$Pr((a\cap h^c)$ = $(a-h)$ = $(a-(a \cap h))$

So, if what I have is true, how can I go for $Pr(a\cap 1-h)$ to $(a-(a \cap h))$?

I was kind of hoping in the solution can be explained, or if any links where I would be able to see that the solution works as well as others so that I can review what I have forgotten over the years.

Thank you.
I will show
[tex] a \cap h^c = a - h [/tex]
first you have to know what
[tex] a - h [/tex] means all elements in "a" except the elements in h ( i.e elements in the intersection of a and h )
I will use ven diagrams in the attachment, the red circle denote to h, yellow denote to a the square denote to the field or omega (Probability space)
in the first square a-h, i remove from "a" the intersection so the half moon resulted
in the second the gray area represent [tex]h^c[/tex] which intersect with a at the half moon. so they are equivalent
there is another way to show that
[tex]a -h = a \cap h^c[/tex]
I prefer this way, we take an element arbitrary in the left side and show that it is in the right side and take an element in the right and show it is in the left as this
let [tex] x \in a - h[/tex] that means [tex] x \in a [/tex] and [tex]x\notin h[/tex] so [tex] x \in h^c [/tex]
and hence [tex] x \in a \cap h^c[/tex]
let [tex] x \in a\cap h^c [/tex]
so [tex] x\in a [/tex] and [tex] x\in h^c [/tex] since x is in the complement of h [tex] x\notin h[/tex]
this give [tex] x \in a-h [/tex]

I have this $Pr(a\cap h^c)$, so i believe that I can say $Pr(a\cap 1-h)$ is that true?
it is false not true since [tex] 1 -h [/tex] is not defined 1 is not a set what is true is
[tex] Pr(h) + Pr(h^c) = 1[/tex]
instead [tex]h^c = \omega - h [/tex]
 

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