The two most common types of errors made by programmers are syntax errors and errors in logic. for a simple language such as basic the number of such errors is usually small. let x denote the number of syntax errors and y the number of errors in logic made on the first run of a basic program. the sample space for x is {0, 1, 2, 3} and the sample space for y is {0, 1, 2, 3, 4, 5}. assume that the joint density for (x, y) is shown in the following table:

x

y

0

1

2

3

0

.400

.100

.020

.005

1

.300

.040

.010

.004

2

.040

.010

.009

.003

3

.009

.008

.007

.003

4

.008

.007

.005

.002

5

.005

.002

.002

.001

(5 points) find the probability that a randomly selected program will have neither of these types of errors.

(5 points) find the probability that a randomly selected program will contain at least one syntax error and at most one error in logic.

(5 points) find the marginal densities for x and y.

(5 points) find the probability that a randomly selected program contains at least two syntax errors.

(5 points) find the probability that a randomly selected program contains one or two errors in logic.

(5 points) are x and y independent?

(10 points) find e[x], e[y], e[xy], and cov(x, y)

(10 points) find var(x), var(y), and rhoxy

(10 points) find the condition density of y given that x = 2, and the conditional mean of y given that x = 2.

7/2

(decimal: 3.5)

welcome ; )

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.