Find the distribution function of the random variable that h

Find the distribution function of the random variable that has the probability distribution f(x) = x/15 for x = 1,2,3,4,5 If the value of the joint probability distribution of X and Y are as shown in the table

Solution

3)

X<1 F(X)=sum(x<=0)f(X)=0

1<=x<2, F(X)=sum(x<=1) f(x)=1/15

2<=x<3 F(X)=f(0)+(1)+f(2)=3/15

3<=x<4 F(X)=7/15

4<=x<5 F(X)=12/15

x>=5 F(X)=1

4. a) P(X=1,Y=2)=1/20 (from the table)

b) P(X=0,Y<3)

to calculate this we look at the probabiities for x=0 and Y=0,1,2 and these are added.

=1/12+1/4+1/120=0.34166

c) P(X+Y<=1)=P(X=0, Y=0)+P(X=0,Y=1)+P(X=1,Y=0)

=1/12+1/4+1/6=0.9166

d)P(X>Y)=P(X=1,Y=0)+P(X=2,Y=0,)P(X=2,Y=1)=1/12+1/6+1/40=0.275