# Department of Mathematics | Illinois State University

```Krzys’ Ostaszewski: http://www.krzysio.net
Author of the ASM Manual (the “Been There Done That!” manual) for Course P/1
http://smartURL.it/krzysioP (paper) or http://smartURL.it/krzysioPe (electronic)
Instructor for Course P/1 online seminar: http://smartURL.it/onlineactuary
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Exercise for June 7, 2008
November 2000 Course 1 Examination, Problem No. 7, Study Note P-09-05,
Problem No. 36
A group insurance policy covers the medical claims of the employees of a small
company. The value, V, of the claims made in one year is described
by V = 100000Y , where Y is a random variable with density function
&quot;\$ k (1 ! y )4 , for 0 &lt; y &lt; 1,
f ( y) = #
otherwise,
%\$ 0,
where k is a constant. What is the conditional probability that V exceeds 40000, given
that V exceeds 10000?
A. 0.08
B. 0.13
C. 0.17
D. 0.20
E. 0.51
Solution.
First we determine k, by using the fact that the integral of a density function must be 1:
1
1
k
k
4
5
1 = ! k (1 &quot; y ) dy = &quot; (1 &quot; y ) = .
5
5
0
0
Therefore k = 5. Now we can find the probabilities needed
Pr (V &gt; 10000 ) = Pr (100000Y &gt; 10000 ) = Pr (Y &gt; 0.1) =
1
=
! 5 (1 &quot; y )
4
dy = &quot; (1 &quot; y )
5 1
0.1
0.10
= 0.9 5 # 0.59,
Pr (V &gt; 40000 ) = Pr (100000Y &gt; 40000 ) = P (Y &gt; 0.4 ) =
1
=
! 5 (1 &quot; y ) dy = &quot; (1 &quot; y )
4
0.4
It now follows that
5 1
0.4
= 0.6 5 = 0.078.
Pr (V &gt; 40000 V &gt; 10000 ) =
=
Pr ({V &gt; 40000} ! {V &gt; 10000})
Pr (V &gt; 10, 000 )
=
Pr (V &gt; 40, 000 ) 0.078
=
= 0.132.
Pr (V &gt; 10, 000 ) 0.590