Ryan Davis
EECS 310 Problem Set 8
1.a) The probability that π(1) = 1 is 1/ S. Since π is a random bijection, π(1) is uniformly distributed over the interval [1,
n] with possible discrete outcomes {1,…, n}. There are S possible outcomes, each that occurs with equal probability, so
the probability of a single given outcome is 1/ S.
b) The probability that f(A) = π(a) is
S  π(a)
C
A1
/
S
C
A
.
The denominator represents the total number of ways to select a subset A of size A from S. We choose A elements from
a set of S elements, the number of ways which to do is simply
S
C
A
. The numerator is significantly less intuitive. It
gives the number of ways to choose a subset A such that π(a) is the minimum element in A. Since we know that π(a) is the
minimum element of A, A must include A  1 other elements that are greater than π(a). The number of elements that are
greater than π(a) is S  π(a), which follows simply from the fact that the original set S has elements {1,…., n}. Thus we
must choose A  1elements from a set of S  π(a), given simply by
S  π(a)
C
A1.
We give an example to illustrate the correctness of the formula. Consider S = {1, 2, 3, 4, 5}, A = 3, and π(a) = 2. The
5
C
3
possibilities for A are
{1, 2, 3}
{1, 2, 4}
{1, 2, 5}
{1, 3, 4}
{1, 3, 5}
{1, 4, 5}
{2, 3, 4}
{2, 3, 5}
{2, 4, 5}
{3, 4, 5}
There are
S  π(a)
C
A1
=
5 – 2
C
2
=
3
C
2
= 3 sets in which π(a) = 2 is the lowest number.
c) We can build on the approach described in b to show that the probability that f(A) = f(B) is
(Σ
J=1,…, S  1
S  J
C
A  1
*
S  J
C
B  1
)
/ (
S
C
A
*
S
C
B
).
The denominator gives the total number of ways to choose a subset of cardinality A and a subset of cardinality B, each
from a set of cardinality S. Assume that A and B both have a smallest element J, so f(A) = f(B). A must have A – 1 other
distinct elements, each of which is greater than J. The total number of elements greater than J is S  J. Thus the number of
ways to choose a subset A in which J is the smallest element is
S  J
C
A  1
. We can perform the same analysis to show that
for the same J, the number of ways to choose a subset A in which J is the smallest element is
S  J
C
B  1
. In the numerator,
we sum over all possible smallest elements J to find the total number of ways that we can choose subsets A and B such
that J is the smallest element in both A and B.
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View Full DocumentWe illustrate the formula with an example in which S = 5, A = 3, and B = 2. The possibilities for A and B are given
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 Fall '11
 N/A
 Economics, Probability, Probability theory, smallest element

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