MA416 Final

Two fair dice are rolled. X is the largest value obtained on any die and Y is the sum of the values. Find the joint probability mass function of X and Y.

The joint probability mass function will be the number of ways a pair of numbers can be arranged over the number of possible arrangement.

The number of possible combinations is 6 × 6 = 36

P(1,2) = 1/36 becuase 1 and 1 is the only combination that can add to 2

P(2,3) = 2/36 becuase 2,1 and 1,2 both add to 3

Suppose that 3 balls are chosen without replacement from an urn consisting of 5 white balls and 8 red balls. Let X_i equal 1 if the ith ball is white, and let it equal 0 otherwise. Give the joint probability mass function of X₁ and X₂.

The joint probability mass function of X₁ and X₂ means the joint pmf of the first two picks. There are 4 pmfs because there are two picks with two options for each pick. The probability is equal to the number of the color of ball out of the total times the new number of the color of ball being considered over the total - 1.

For example: For the first and second pick being red, p(0,0) = (8/13) x (7/12)

Thre bowls are labeled 1, 2, and 3. Bowl i contains i white balls and 5 - i red balls. In an experiment, a bowl is randomly selected, and then 3 balls are randomly selected from that bowl without replacement.

a. Given that bowl 1 is NOT selected, what is the probability of drawing exactly 2 red balls?

b. What is the probability that exactly 2 red balls are drawn?

c. Given that exactly two red balls were drawn, what is the probability that bowl 3 was selected?

The probability that a number of red balls is chosen from any bowl is:

(total number of red balls choose specified number of red balls)(total number of white balls choose specified number of white balls) / (total number of balls choose specific number of balls chosen)

a. 0.5P(2 red balls are chosen from bowl 2) + 0.5P(2 red balls are chosen from bowl 3)

b. 0.333P(2 red balls are chosen from bowl 1) + 0.333P(2 red balls are chosen from bowl 2) + 0.333P(2 red balls are chosen from bowl 3)

c. P(bowl 3 given 2 red) = P(bowl 3 and 2 red) / P(2 red) = P(2 red given bowl 3)P(bowl 3) / P(2 red)

What is the formula for conditional probability? P(A|B)

P(A given B) = P(A and B) / P(B) = P(B given A)P(A) / P(B)

A company sells CDs that are defective with a probability of 0.05. The company sells the CDs in packages of 10, with a money back garuntee that only at most 1 of the CDs per package will be defective.

a. What is the probability that a package is returnable for the money back garuntee?

b. If someone buys 4 packages, what is the probability that they will be able to return 2 of the packages for the money back garuntee?

c. The cost, C, to the manufacturer is given by C = Y² + 5Y + 1, where Y is the number of returnable packages in a shipment. What is the expected cost for a shipment of 100 packages?

a. This is asking for P(X>1) = 1 - P(X<=1) = 1 - (P(X=0) + P(X=1))

b. P(A of B packages returnable) = (B A)P(package is returnable)^{number of packages returning (A)}P(package is not returnable)^{numbrt of pakages not returning (B - A)}

c. This is a binomial distribution Bin(100, probability a package can be returned)

E[C] = E[Y² + 5Y + 1] = E[Y²] + 5E[Y] +1

E[Y] = np

E[Y²] = Var(Y) + E[Y]²

Var(Y) = np(1 - p)

What is the equation for E[X²]?

E[X²] = Var(X) + E[X]²

What is the equation for the mean of a binomial distribution, E[X]?

E[X] = np where Bin(n, p) where n is the number of things and p is the probability for one of those things

What is the equation for the variance of a binomial distribution, Var(X)?

Var(X) = np(1-p) where Bin(n, p) where n is the number of things and p is the probability for one of those things

What is the standard equation for variance, Var(X)?

What is the standard equation for the expected value, E[X]?

Suppose that in two neighboring counties, road repairs per weem are considered “major” is the amount of roads under repair is 3 miles or longer. Repair lengths in both counties range from 0 to 4 miles long. X is the length of repair in one county, and Y is the length of repair in a second county in one week. The joint PDF of X and Y is given by f(x,y) = xy/64 for 0<=x,y<=4.

a.

b. Show that X and Y are independent and have identical distributions, providing the marginal PDF they share.

The 2 counties want to hire a single company for repairs. One company will only handle combined jobs of a maximum of 6 miles at a time for a given week before charging huge additional fees. Use a probabilistic arguement to recommend whether or not the company should be hired for the repairs.

a.

b. 1. f_X(x) = intergral over bounds of joint PDF dy

2. f_Y(y) = integral over bounds of joint PDF dx

3. f_XY = joint PDF = f_Xf_Y = (first integral)(second integral) <=> X perpendicular Y

c. Draw the square from 0 to X and from 0 to Y on a graph. 6 miles a week means x+y=6 => y=-x+6. Draw this line intersecting the square. Double integrate the PDF from the X and Y bounds of the rectangular portion, and add to the double integration from X to Y of the portion under the line y=-x+6. This number should be small enough that it is not probable that the huge fee is incurred.

X and Y are random variables with Var(X)=8 and Var(Y)=6.

a. If X and Y are independent, what is the variance of 6X+3Y+2?

b. If X and Y have correlation 0.4, what is the variance of X-2Y?

a. Var(aX+bY) = a²Var(X) + b²Var(Y)

b. Var(aX+bY) = a²Var(X) + b²Var(Y) - abCov(X,Y) where Cov(X,Y) = Corr(X,Y)(sqrt Var(X))(sqrt Var(Y))

What is the equation for covariance?

Cov(X,Y) = Corr(X,Y)(sqrt Var(X))(sqrt Var(Y))

What is the equation for variance with correlation?

Var(aX+bY) = a²Var(X) + b²Var(Y) - abCov(X,Y)

X and Y are random variables with joint PDF f(x,y) = x + y for 0<x,y<1 and 0 otherwise.

a. What is the integral that finds the probability that X is greater than 2Y?

b. Find the conditional distribution of Y given X.

c. Find the probability that Y is greater than 0.25 given that X is 0.5.

d. Find the expected value of Ygiven that X is 0.75.

a. The PDF is x+y, but the line that describes the probability that is being asked for is X=2y → Y=0.5X, so P(X>2Y) = integral from 0 to 1 integral from 0 to 0.5x of x+y dy dx

b. f_Y|X = f_X,Y / f_X

f_Y,X = x+y

f_X = integral from 0 to 1 of x+y dy

c. Use f_Y|X found above and set X=0.5

P = integral from 0.25 to 1 of f_Y|X=0.5 dy

d. Use f_Y|X found above and set X=0.75

P = integral from 0 to 1 of y times f_Y|X=0.75 dy

What is the integral that solves for expected value?

f(y) = cy²(1-y²) for -1<y<1 and f(y) = 0 elsewhere

a. Find c such that f(y) is a density function.

b. Find the mean and variance of Y.

a. 1 = integral over bounds of f(y) dy - solve for c

b. E[Y] = integral over bounds of y x f(y) dy = 0

What is the integral for a probability density function?

What is the intergal for a probability mass function?