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Jim Lambers Math 6A Spring Quarter 2003- Practice Final Exam
- Suppose that you are rolling a six-sided die five times. What is the probability that the first roll is a 1, the second roll is a 2, the third roll is a 3, and the fourth roll is a 4?
- Suppose that you are rolling a six-sided die five times. What is the probability that the first roll is not a 1, the second roll is not a 2, the third roll is not a 3, and the fourth roll is not a 4?
- Suppose that you are flipping a fair coin three times. Let E 1 be the event that the the first flip comes up heads, and let E 2 be the event that two of the three flips are heads. Determine whether E 1 and E 2 are independent. Note: simply writing down whether they are independent or not is not sufficient for full credit.
- Suppose that you are rolling a six-sided die 10 times. What is the probability that exactly 4 of the rolls are a 6? A formula for the answer is sufficient; you do not need to carry out all of the arithmetic.
- Consider the statement “If you did the homework problems, then you will do well on this exam. If you do well on this exam, you will get a good grade in this course.” Using the rules of inference, what is a valid conclusion that can be made from this statement?
- Why is the following argument invalid?
“If I get a ticket, then I was driving too fast. I did not get a ticket, so therefore I was not driving too fast.”
- Prove by induction that ∑n
i=
(i − 1)^2 = n(n − 1)(2n − 1) 6 for each positive integer n.
- Write down the generating function for the geometric progression whose first few terms are
3 , 2 , 4 / 3 , 8 / 9 , 16 / 27.
- Write down a recursive definition for the function f : N → N, f (n) = 2n(n+1)/^2 , n ≥ 0. Your definition must include a basis step and a recursive step.
- Given the following recursive definition for a complete binary tree:
- basis step: a single vertex is a complete binary tree.
- recursive step: if T 1 and T 2 are complete binary trees of the same height, and r is a vertex, then the tree T 1 · T 2 formed by adding edges between r and the roots of T 1 and T 2 is a complete binary tree.
use structural induction to prove that if T is a complete binary tree of height h, then T has n vertices, where n = 2h+1^ − 1.
- Solve the recurrence relation
an = 6an− 1 − 8 an− 2 , a 0 = 2, a 1 = 3.
- Solve the nonhomogeneous recurrence relation
an = 6an− 1 − 8 an− 2 + 3n, a 0 = − 1 , a 1 = 2.
- Assume that a university has 5,431 students. Of these students, 1,712 have taken a calculus course, 435 have taken a discrete mathematics course, and 1,024 have taken a French course. Furthermore, 218 have taken both calculus and discrete mathematics, 721 have taken both calculus and French, 171 have taken both discrete mathematics and French, and 35 have taken all three courses. How many students at this university have not taken any of the three courses?
- Find the generating function for the sequence that solves the recurrence relation
an = 2an− 1 + n + 1, a 0 = 1.
You do not need to find the sequence {an}, only its generating function.
Possibly Useful Formulas
E(X) =
s∈S
X(s)p(s)
E(X) =
r∈X(S)
rp(X = r)
E(X + Y ) = E(X) + E(Y )
V (X) =
s∈S
(X(s) − E(X))^2 p(s)
V (X) = E(X^2 ) − E(X)^2 Rules of Inference Tautology Name p → (p ∨ q) Addition (p ∧ q) → p Simplification ((p) ∧ (q)) → (p ∧ q) Conjunction [p ∧ (p → q)] → q Modus ponens [¬q ∧ (p → q)] → ¬p Modus tollens [(p → q) ∧ (q → r)] → (p → r) Hypothetical syllogism [(p ∨ q) ∧ ¬p] → q Disjunctive syllogism [(p ∨ q) ∧ (¬p ∧ r)] → (q ∨ r) Resolution ∀xP (x) → P (c) Universal instantiation P (c) for an arbitrary c → ∀xP (x) Universal generalization ∃xP (x) → P (c) for some element c Existential instantiation P (c) for some element c → ∃xP (x) Existential generalization Fallacies Contingency Name [(p → q) ∧ q] → p Affirming the conclusion [(p → q) ∧ ¬p] → ¬q Denying the hypothesis [p ↔ q] → q Circular reasoning
∑^ n
k=
n k
= 2n
∑n
k=
k
n k
= n 2 n−^1
∑^ n
k=
rk^ = rn+1^ − 1 r − 1
, r 6 = 1
