Mathematics, 16.10.2019 05:10 nevaeh33cx

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 ?

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Answer from: minideeri

A. c=1/100

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

C. $f_{x}(x)=\frac{x+1}{10}\ y\ f_{y}(y)=\frac{y+1}{10}$

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 $f_{xy}\$ is a probability mass function, then (for a discrete variables because that´s our case)

$\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{f_{xy}}}=1\\$

So, for our problem we have:

$\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{c(x_{i}+1)(y_{j}+1)}}=1\\$

And we solve for c:

$c\sum_{i=1}^{\infty}{\sum_{j=1}^{\infty}{(x_{i}+1)(y_{j}+1)}}=1\\$

$c\sum_{i=1}^{\infty}{(x_{i}+1)\sum_{j=1}^{\infty}{(y_{j}+1)}}=1\\$

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

$\sum_{i=1}^{\infty}{(x_{i}+1)=(0+1)+(1+1)+(2+1)+(3+1)=10$

And we use that to solve for c:

$c\sum_{i=1}^{\infty}{(x_{i}+1)(10)=1\\$

$c(10)(10)=1\\$

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:

$f_{x}(x)=\sum_{j=1}^{\infty}{\frac{(x+1)(y_{j}+1)}{100}$

$f_{x}(x)=\frac{10(x+1)}{100}$

$f_{x}(x)=\frac{1}{10}(x_{i}+1)\\$

And we do the same for y:

$f_{y}(y)=\frac{1}{10}(y_{i}+1)\\$

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

Answer from: Quest

step-by-step explanation:

Answer from: Quest

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

$\frac{ab}{bc} =\frac{ad}{ab}$

now, we can cross multiply

and we get

$ab\timesab=ad\times bc$

$ab^2=ad\times bc$

so, this could be step to prove it

Seth is using the figure shown below to prove pythagorean theorem using triangle similarity: in the

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Consider two discrete random variables x, y , taking values in {0, 1, 2, 3} each (for a total of 16...

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