Let's calculate those joint probabilities:

e.g. #P(X=2 wedge Y=3)=0.15625#.

It is now quite easy to obtain separate probability distributions of #X# and #Y#: we should sum rows or columns of area B2:E3, respectively, e.g.

#P(X=2)=sum_(i=1)^4 P(X=2 wedge Y=i)=0.09375+ldots+0.1875=0.5625#

Now we can use known formulas for the values we're looking for:

#mu_x="E"X=1*0.4375+2*0.5625=1.5625#

#mu_y="E"Y=1*0.15625+2*0.21875+3*0.28125+4*0.34375=2.8125#

#"E"X^2=1^2*0.4375+2^2*0.5625=2.6875#

#"E"Y^2=1^2*0.15625+ldots+4^2*0.34375=9.0625#

#sigma_x^2="E"X^2-"E"^2X=2.6875-1.5625^2=0.24609375#

#sigma_y^2="E"Y^2-"E"^2Y=9.0625-2.8125^2=1.15234375#

Now for the correlation coefficient #rho# we need the covariance #sigma_(xy)# so let's find the probability distribution of #XY#: (here #p_(ij)=P(X=i wedge Y=j)# )

#P(XY=1)=p_(11)=0.0625#

#P(XY=2)=p_(12)+p_(21)=0.1875#

#P(XY=3)=p_(13)=0.125#

#P(XY=4)=p_(14)+p_(22)=0.28125#

#P(XY=5)=0#

#P(XY=6)=p_(23)=0.15625#

#P(XY=7)=0#

#P(XY=8)=p_(24)=0.1875#

Now, for the covariance:

#"E"(XY)=1*0.0625+ldots+8*0.1875=4.375#

#sigma_(xy)="E"(XY)-"E"X*"E"Y=-0.01953125#

#rho=(sigma_(xy))/sqrt(sigma_x^2*sigma_y^2)=-0.0366765706779718#