CS 1050 B: Constructing Proofs
April 2, 2008
Solutions to Homework 6
Lecturer: Sasha Boldyreva
Problem 6.1, 20 points.
Problem 8 from Section 6.2 of Rosen’s textbook.
a) Since 1 has either t precede 2 or follow it, and there is no reason that one of these should
be any more likely than the other, we immediately see that the answer is 1/2.
b) Similarly, the answer is 1/2.
c) For 1 to immediately precede 2, think of the two numbers glued together. Then we are
really permuting n1 numbers – the single numbers from 3 to n and the glued number, 12.
There are (
n

1)! ways to do this. And there are
n
! permutations total. Hence the answer
is (
n

1)!
/n
!.
d) Half of the permutations have
n
preceding 1. Of these permutations, half of them have
n1 preceding 2. Therefore one fourth of the permutations satisfy these conditions, so the
probability is 1/4.
e) Looking at the relative placements of 1,2, and n, we see that one third of the time, n will
come first. Thus the answer is 1/3.
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 Spring '05
 HUANG
 Conditional Probability, Probability theory, Bayesian probability

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