Answer the last three . Among married couples in a certain community, let X be the number of previous marriages for thewife and Y be the number previous marriages for the husband, with joint probability mass function120.30 0.10 |0.050.15 0.15 0.050.05 0.05 0.1012If a couple in the community is randomly selected, what is the probability the husband is onhis first marriage?а.b.What is the mean umber of previous marriages for men in the community?с.What is the standard deviation for the number of previous marriages for men in thecommunity?d.Are X and Y independent?Compute Cov(X, Y).е.

c) The probability distribution for variable Y is, Y 0 1 2 P(Y) 0.3+0.15+0.05=0.5 0.1+0.15+0.05=0.3 0.05+0.05+0.1=0.2 Men number of previous marriages for men in the community is, E(Y)=∑YP(Y) =(0*0.5)+(1*0.3)+(2*0.2) =0+0.3+0.4 =0.7 Standard deviation for the number of previous marriages for men in the community is, SD(Y)=E(Y2)−[E(Y)]2 =(0*0.5)+(12*0.3)+(22*0.2)−0.72 =0+0.3+0.8−0.49 =0.7810d) Condition for Independence: If X and Y are random variables, then P(X,Y)=P(X)*P(Y). Test the condition for X=0, Y=0 From the given information, X 0 1 2 P(X) 0.3+0.10+0.05=0.45 0.15+0.15+0.05=0.35 0.05+0.05+0.10=0.2 Consider, E(X)=∑XP(X) =(0*0.45)+(1*0.35)+(2*0.2) =0+0.35+0.4 =0.75 Y 0 1 2 P(Y) 0.3+0.15+0.05=0.5 0.1+0.15+0.05=0.3 0.05+0.05+0.1=0.2 Consider, P(X=0,Y=0)=0.30 P(X=0)=0.45 P(Y=0)0.5 P(Y=0)*P(X=0)=0.5*0.45 =0.225 ≠P(X=0,Y=0) That is, the condition P(X,Y)=P(X)*P(Y) is not satisfied. Hence, the variables X and Y are not independent. e) Cov(X,Y): Cov(X,Y)=E(XY)-E(X)E(Y) =∑XYP(X,Y)-E(X)E(Y) =0*0*0.30+0*1*0.10+0*2*0.05+1*0*0.15+1*1*0.15+1*2*0.05+2*0*0.05+2*1*0.05+2*2*0.10-0.7*0.75 =0.75-0.525 =0.225