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Exercise for January 19, 2008
May 1982 Course 110 Examination, Problem No. 18
Two balls are dropped in such a way that each ball is equally likely to fall into any one of
four holes. Both balls may fall into the same hole. Let X denote the number of
unoccupied holes at the end of the experiment. What is the moment generating function
of X?
A.
7 1
9
21
B. t + t 2
! t if t = 2 or 3, otherwise
4 2
4
8
t
$
1
1 2t
1 ! 3t
C.
D.
E. # e 4 + 3e 4 &
3e2t + e3t
e + 3e3t
4
4
4"
%
(
)
(
)
Solution.
Each ball had four possible holes, so that the total number of ways for the balls to fall is
4 ! 4 = 16. There can be only two or three holes unoccupied. If three holes are
unoccupied, there are four ways to put two balls in one occupied hole. Hence,
4 1
Pr ( X = 3) =
= ,
16 4
3
Pr ( X = 2 ) = 1 ! Pr ( X = 3) = .
4
Therefore,
3
1
1
M X ( t ) = e2t + e3t = 3e2t + e3t .
4
4
4
Answer C.
(
)
© Copyright 2004-2008 by Krzysztof Ostaszewski.
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Exercises from the past actuarial examinations are copyrighted by the Society of
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