Consider two discrete random variables x, y , taking values in {0, 1, 2, 3} each (for a total of 16 possible points). their joint probability mass function is given by fxy (x, y) = c · x+1 y+1 . answer the following questions. (a) what is c? (b) what is the probability that x = y ? (c) derive the marginal probability mass functions for both x and y . (d) what is the probability that x ≤ y given that y = 2? (e) what is the probability that x ≤ y ?

A. c=1/100

B. P(x==y)=3/10

C.

D. P(x≤y | y=2)=9/50

E. P(x≤y)= 7/20

Step-by-step explanation:

A. To answer the question we use a really important property about mass functions of probability: If is a probability mass function, then (for a discrete variables because that´s our case)

So, for our problem we have:

And we solve for c:

Because x and y only takes values in [0,1,2,3] we know:

And we use that to solve for c:

Finally we have c=1/100

B. For part B, we just choose the probabilities that we want because there are few values to choose and use the mass function to find them:

P(x=0,y=0)=(0+1)(0+1)/100=1/100

P(x=1,y=1)=4/100 (putting x=1 and y=1 in the function)

P(x=2,y=2)=9/100

P(x=3,y=3)=16/100

And for the last part we add them:

P(X==Y)=1/100+4/100+9/100+16/100=3/10

This is the answer

C. We find (in discrete variables) the marginal probability mass function just solving the function for one of the variables as follows:

And we do the same for y:

These functions are the marginal probability mass functions for x and y

D. Because there are few values of x and y that they can be. we choose the probabilities as follows:

P(x=0,y=2)=(0+1)(1+2)/100=3/100

P(x=1,y=2)=6/100

P(x=2,y=2)=9/100

And we add them to find our answer

P(x≤y | y=2)=3/100+6/100+9/100=9/50

E. We are working with a few amount of possible values, so we can use the part B (P(x==y)) and add them the probabilities that are left to complete P(x≤y)

P(x==y)=3/10 (B)

P(x=0,y=1)=2/100

P(x=0,y=2)=3/100

P(x=1,y=2)=6/100

P(x=0,y=3)=4/100

P(x=1,y=3)=8/100

P(x=2,y=3)=12/100

We add them to get our final

P(x≤y)=7/20

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.

step-by-step explanation:

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.