Math 213, Section X1 Midterm Exam 2 Solutions Spring 2006
Problem 1
Suppose you roll 4 ordinary dice. Determine the following probabilities:
1. The probability that at least one 6 comes up.
Solution: By the complement trick, this is 1 −P(no 6 in 4 rolls), i.e., 1 −(5/6)4.
2. The probability that exactly one 6 comes up.
Solution: This is given by the binomial distribution: 4
1(1/6)1(1 −1/6)3=4·53
64.
3. The probability that all four numbers showing up are distinct.
Solution: This is a birthday-type probability: 6·5·4·3
64=5
18 .
4. The probability that all four dice show the same number.
Solution: There are 6 ways to choose the number showing up on the dice, and for a
given number, the probability that all four dice show this particular number is 1/64.
Thus, the probability in question is 6 ·(1/64) = 1/63.
Problem 2
Consider the recurrence an=−2nan−1.
1. (Multiple choice. Circle the correct answer.) Determine whether this recurrence is
(a) Linear and homogeneous
(b) Linear and nonhomogeneous
(c) Nonlinear and homogeneous
(d) Nonlinear and nonhomogeneous
Solution: (a) The recurrence is linear and homogeneous. (Note that in a linear recur-
rence the coefficients of the terms an,an−1, etc., don’t have to be constant and can be
arbitrary functions of n.)
2. Solve this recurrence with initial condition a0= 1.
Solution: By iteration,
an=−2nan−1= (−2)2n(n−1)an−2=··· = (−2)nn(n−1) . . . 2·1a0= (−2)nn!.
Problem 3
Let andenote the number of ways to climb nstairs if stairs can be climbed 2 or 3 at a time.
Set up a recurrence for an; be sure to clearly explain how each term in this relation arises.
Solution: Consider the last step when climbing nstairs. If this is a 2-stair step, your position
before taking this step is at the end of n−2 stairs, and there are an−2ways to reach this
position. If it is a 3-stair step, your position before taking the step is at the end of n−3
stairs, and there are an−3ways to reach this position. Since these are the only possibilities
for the last step, the sum of these two counts gives the total number of ways to climb nstairs,
i.e., we have the recurrence an=an−2+an−3.
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