Actuarial Outpost Conditional Probability
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#1
11-16-2015, 07:49 PM
 tvossler Member SOA Join Date: Sep 2014 Location: Nevada Studying for Exam p College: UoP--Graduated Favorite beer: Bushmills Irish Whiskey Posts: 62
Conditional Probability

In general if P(A)>0, then P(B|A) = P(A and B)/P(A).

Sometimes, as in the case of SOA #17 from November 2001: The loss due to a fire in a commercial building is modeled by a random variable X with density function f(x) =.005(20-x) for 0<x<20

Given that a fire loss exceeds 8, what is the probability that it exceeds 16?

That conditional probability is shortened to P(B|A) =P(AandB)/P(A) = P(B)/P(A).

What is this due to? Why does it sometimes get reduced to this and other times not?

Thank you.
#2
11-16-2015, 08:00 PM
 therealsylvos Member CAS Join Date: Mar 2014 Posts: 14,483

When P(A and B) = P(B). If the loss is 16, it's certainly greater than 8. So P(Greater than 16 and greater than 8) = P(Greater than 16).
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#3
11-16-2015, 08:30 PM
 tvossler Member SOA Join Date: Sep 2014 Location: Nevada Studying for Exam p College: UoP--Graduated Favorite beer: Bushmills Irish Whiskey Posts: 62

Thank you very much.
#4
11-16-2015, 09:26 PM
 Abelian Grape Meme-ber                         Meme-ber CAS Join Date: Jul 2014 Posts: 38,746

np
#5
11-17-2015, 01:50 AM
 KyleMoyer Member SOA Join Date: Sep 2015 College: West Chester recent graduate Posts: 39

The best way to do it is write it out. If you write out "X > 8 and X > 16," you realize it looks kinda silly as opposed to (for example) "X > 8 and X < 16."

Granted, that's pretty straightforward, but you can get something a bit more involved where you write it out and realize it's redundant
#6
11-17-2017, 05:47 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 336

If $B \subseteq A$ then $A \cap B=B$.

Hence $P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{P(B)}{P(A)}$

I believe this is really common on exam problems.

 Tags conditional probability, soa #17 november 2001

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