Solve the following 1. If the joint probability distribution of X

slaggingV

slaggingV

Answered question

2021-09-20

The following problem
1. If X and Y's combined probability distribution is given by
f(x,y)=x+y30 for x=0,2,3y=0,1,2 
Find 
a. P(X2,Y=1) 
b. P(X>2,Y1) 
c. P(X>Y)

Answer & Explanation

saiyansruleA

saiyansruleA

Skilled2021-09-21Added 110 answers

Here,
xyf(x,y)000200.066667300.1010.033333210.1310.133333020.066667220.133333320.166667
a)
P(X2,Y=1)=P(X=0,Y=1)+P(X=2,Y=1)
=f(0,1)+f(2,1)
=0.33333+0.1
=0.133333
b)
P(X>2,Y1)=P(X=3,Y=0)+P(X=3,Y=1)
f(3,0)+f(3,1)
=0.1+0.133333
=0.233333
c)
P(X>Y)=P(XY>0)=P(X=2,Y=0)+P(X=3,Y=0)+P(X=2,Y=1)+P(X=3,Y=1)+P(X=3,Y=2)
=0.066667+0.1+0.1+0.133333+0.166667
=0.566667

user_27qwe

user_27qwe

Skilled2023-06-17Added 375 answers

Result:
a. P(X2, Y=1) = 415.
b. P(X>2, Y1) = 730.
c. P(X>Y) = 715.
Solution:
a. P(X2, Y=1):
To calculate this probability, we need to sum up the probabilities for all (X, Y) pairs that satisfy the condition X2 and Y=1.
We have three possible pairs: (0, 1), (2, 1), and (3, 1). Therefore, we can write the probability as follows:
P(X2, Y=1) = P((X=0, Y=1) (X=2, Y=1) (X=3, Y=1))
To calculate the probability for each pair, we substitute the corresponding values into the PDF:
P(X=0, Y=1) = f(0, 1) = 0+130 = 130
P(X=2, Y=1) = f(2, 1) = 2+130 = 330 = 110
P(X=3, Y=1) = f(3, 1) = 3+130 = 430 = 215
Now, we can sum up these probabilities to get the final result:
P(X2, Y=1) = 130 + 110 + 215 = 130 + 330 + 430 = 830 = 415.
Therefore, P(X2, Y=1) = 415.
b. P(X>2, Y1):
Similar to part a, we'll sum up the probabilities for all (X, Y) pairs that satisfy the condition X>2 and Y1.
The possible pairs that satisfy this condition are: (3, 0) and (3, 1).
P(X>2, Y1) = P((X=3, Y=0) (X=3, Y=1))
Calculating the probability for each pair:
P(X=3, Y=0) = f(3, 0) = 3+030 = 330 = 110
P(X=3, Y=1) = f(3, 1) = 3+130 = 430 = 215
Summing up these probabilities:
P(X>2, Y1) = 110 + 215 = 330 + 430 = 730.
Therefore, P(X>2, Y1) = 730.
c. P(X>Y):
To calculate this probability, we need to sum up the probabilities for all (X, Y) pairs where X>Y.
The pairs that satisfy this condition are: (2, 0), (3, 0), (3, 1), and (3, 2).
P(X
>Y) = P((X=2, Y=0) (X=3, Y=0) (X=3, Y=1) (X=3, Y=2))
Calculating the probability for each pair:
P(X=2, Y=0) = f(2, 0) = 2+030 = 230 = 115
P(X=3, Y=0) = f(3, 0) = 3+030 = 330 = 110
P(X=3, Y=1) = f(3, 1) = 3+130 = 430 = 215
P(X=3, Y=2) = f(3, 2) = 3+230 = 530 = 16
Summing up these probabilities:
P(X>Y) = 115 + 110 + 215 + 16 = 230 + 330 + 430 + 530 = 1430 = 715.
Therefore, P(X>Y) = 715.
karton

karton

Expert2023-06-17Added 613 answers

Step 1:
a. To find P(X2,Y=1), we need to sum up the probabilities for all possible combinations of X and Y that satisfy this condition. From the given probability distribution, we have:
P(X2,Y=1)=P(X=0,Y=1)+P(X=2,Y=1)=0+130+2+130=430=215
Therefore, P(X2,Y=1)=215.
Step 2:
b. To find P(X>2,Y1), we again need to sum up the probabilities for all possible combinations that satisfy this condition:
P(X>2,Y1)=P(X=3,Y=0)+P(X=3,Y=1)=3+030+3+130=730
Hence, P(X>2,Y1)=730.
Step 3:
c. Finally, to find P(X>Y), we need to sum up the probabilities for all combinations where X is greater than Y:
P(X>Y)=P(X=2,Y=0)+P(X=3,Y=0)+P(X=3,Y=1)=2+030+3+030+3+130=930=310
Thus, P(X>Y)=310.
star233

star233

Skilled2023-06-17Added 403 answers

To solve the given problem, we need to calculate the probabilities using the given probability distribution function (PDF). Let's calculate each part step by step:
a. To find P(X2,Y=1), we need to sum up the probabilities for all pairs (x, y) that satisfy the condition. So we have:
P(X2,Y=1)=f(0,1)+f(2,1)+f(3,1)
Substituting the values from the PDF, we get:
P(X2,Y=1)=0+130+2+130+3+130
Simplifying the expression, we have:
P(X2,Y=1)=130+330+430=830=415
Therefore, P(X2,Y=1)=415.
b. To find P(X>2,Y1), we again need to sum up the probabilities for all pairs (x, y) that satisfy the condition. So we have:
P(X>2,Y1)=f(3,0)+f(3,1)
Substituting the values from the PDF, we get:
P(X>2,Y1)=3+030+3+130
Simplifying the expression, we have:
P(X>2,Y1)=330+430=730
Therefore, P(X>2,Y1)=730.
c. To find P(X>Y), we need to sum up the probabilities for all pairs (x, y) where x is greater than y. So we have:
P(X>Y)=f(2,0)+f(3,0)+f(3,1)+f(3,2)
Substituting the values from the PDF, we get:
P(X>Y)=2+030+3+030+3+130+3+230
Simplifying the expression, we have:
P(X>Y)=230+330+430+530=1430=715
Therefore, P(X>Y)=715.
In summary:
a. P(X2,Y=1)=415
b. P(X>2,Y1)=730
c. P(X>Y)=715

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