Solve the following: if the join probability distribution of X and Y is given by f(x,y)=\frac{x+y}{30}

UkusakazaL

UkusakazaL

Answered question

2021-08-04

Solve the following
If the join probability distribution of X and Y iss given by
f(x,y)=x+y30
f or x=0, 2, 3:
y=0, 1, 2
Find
a) P(X2, Y=1).
b) P(X>2, Y1).
c) P(X>Y).

Answer & Explanation

Ayaana Buck

Ayaana Buck

Beginner2021-08-04Added 2 answers

Step 1
Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.
Here,
xyf(x,y)000200.066667300.1010.033333210.1310.133333020.066667220.133333320.166667
Step 2
a) P(X2, Y=1)=P(X=0, Y=1)+P(X=2, Y=1)
=f(0,1)+f(2,1)
=0.033333+0.1
=0.133333
Step 3
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
Step 4
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

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