Question text is in black, solutions in blue.
Q1: 20 points Q2: 15 points Q3: 15 points Q4: 20 points Q5: 15 points Q6: 20 points Q7: 20 points Total: 125 points
A proof by contradiction is an argument where you first assume the negation of the statement you are trying to prove, then use that assumption and known facts to derive an impossible conclusion.
The intersection of sets A and B is the set of all elements that are in both A and B.
A bijection from A to B is a function f from A to B where every element of B is equal to f(x) for exactly one element x of A.
A permutation from A is a sequence of elements of A, where no element appears more than once in the sequence. P(n, r) is the number of such permutations of length r, if n is the size of A.
The expected value of a random variable is a weighted average of all of its possible values, with each value weighted by its probability. It is the sum, over all possible values v, of v times the probability of v.
If we make a truth table for this proposition, we have eight lines for the eight possible settings of p, q, and r. In only two of them, for "p ∧ ¬q ∧ r" and "p ∧ ¬q ∧ ¬r", is the statement "p → q" false. With three fair coin tosses, the eight lines are each equally likely to describe the true situation, and the statement is true in six of the eight, for a probability of 6/8 or 3/4.
This compound proposition is false in the two situations given above, and also in the two lines where "q → r" is false, which are "p ∧ q ∧ ¬r" and "¬p ∧ q ∧ ¬r". So it is true in four of the eight cases, for a probability of 4/8 or 1/2.
Let P(n) be the statement "if |X| = n, there are exactly 3n
functions from X to Y".
Base Case: n = 0, so X is empty. There is one function from
∅ to Y, the empty relation, and 30 is 1.
Alternate Base Case: n = 1, so X = {a}. There are exactly three
functions from X to Y, since f(a) could be any of the three
elements of Y. And 31 is 3.
Inductive Step: We assume P(n) and want to prove P(n+1). Let X
be a set with n+1 elements. Let Z be X with one element (which
we'll call a) removed. Z has n elements, so by
the inductive hypothesis, there are exactly 3n functions
from Z to Y. To make a function from X to Y, we choose a function
f from Z to Y and choose a value for f(a) to extend this function
to X. There are 3n choices for f, and 3 choices for
f(a), so by the Product Rule for counting, there are 3n
× 3 = 3n+1 functions from X to Y. This is just
what we need to prove for the inductive step.
There are C(52, 3) three-element subsets of a 52-element set. We can compute C(52, 3) as 52 × 51 × 50 / 1 × 2 × 3, or 52 × 17 × 25 = 22100.
There are C(4, 3) = 4 ways to pick three of the four kings, for example. Thus there are four three-of-a-kind sets for each rank, or 4 × 13 = 52 in all.
There are C(13, 3) = 13 × 12 × 11 / 1 × 2 × 3 = 13 × 2 × 11 = 286 ways to pick three of the 13 hearts, for example. The total number of "flush" hands is thus 4 × 286 = 1144, 286 for each suit. The probability of getting a flush is thus 1144/22100 = 22/425.
We will show for all positive integers n that 1 - p(n) =
(5/6)n-1.
Base case: n = 1, p(n) = 0 because I can never have the same
number twice in a row if there is only one throw. And 1 - 0 =
(5/6)1-1.
Inductive Step: We assume that 1 - p(n) = (5/6)n-1
and try to prove that p(n+1) = (5/6)n+1-1 =
(5/6)n. If we throw the die n+1 times, there are two
ways we can get the same number twice in a row -- it could happen
during the first n throws or the last two throws could be the
same. The probability that neither of these things happen (which
is 1 - p(n+1)) is the probability of it not happening in the
first n throws (which is (5/6)n-1) time the
probability that the last throw will be different from the next
to last throw, which is 5/6. So 1 - p(n+1) = (5/6)n-1
× 5/6 = (5/6)n, which is what we needed to prove.
0 1 1
1 1 0
0 2 0
(x)--------->(y)<--\
| <---------^^ \__/
| //
| //
| //
| //
| //
V//
(z)
1 3 0
1 2 1
2 2 0
We want the (y, y) entry of the matrix A4, which we get by taking the dot product of the y row of A2 and the y column of A2. This is (1×3) + (2×2) + (1×2) = 3 + 4 + 2 = 9.
0 1/2 1/2
1/2 1/2 0
0 1 0
1/4 3/4 0
1/4 1/2 1/4
1/2 1/2 0
By the Markov Process theorem, this probability is the (red, red) entry of T2, which is 0 by part (b). The light will be yellow after one minute, and cannot go directly from yellow to red.
By the Markov Process theorem, this probability is the (red, red) entry of T3. We can compute this by taking the dot product of the red (last) row of T and the red (last) column of T2, which is 0×0 + 1×1/4 + 0×0 = 1/4. The only way this can happen is by going from red to yellow (probability 1), then to green (probability 1/2), then to red (probability 1/2).
Last modified 2 December 2013