8.2 Problems on random vectors and joint distributions

Let X be the number of aces and Y be the number of spades. Define the events $A{S}_i$ , A _i , S _i , and N _i , $i=1,2$ , of drawing ace of spades, other ace, spade (other than the ace), and neither on the i selection. Let $P(i,k)=P(X=i,Y=k)$ .

${\displaystyle P(0,0)=P\left(N_1{N}_2\right)=\frac{36}{52}•\frac{35}{51}=\frac{1260}{2652}}$

${\displaystyle P(0,1)=P(N_1{S}_2\bigvee S_1{N}_2)=\frac{36}{52}•\frac{12}{51}+\frac{12}{52}•\frac{36}{51}=\frac{864}{2652}}$

${\displaystyle P(0,2)=P\left(S_1{S}_2\right)=\frac{12}{52}•\frac{11}{51}=\frac{132}{2652}}$

${\displaystyle P(1,0)=P(A_1{N}_2\bigvee N_1{S}_2)=\frac{3}{52}•\frac{36}{51}+\frac{36}{52}•\frac{3}{51}=\frac{216}{2652}}$

${\displaystyle P(1,1)=P(A_1{S}_2\bigvee S_1{A}_2\bigvee A{S}_1{N}_2\bigvee N_{1}A{S}_2)=\frac{3}{52}•\frac{12}{51}+\frac{12}{52}•\frac{3}{51}+\frac{1}{52}•\frac{36}{51}+\frac{36}{52}•\frac{1}{51}=\frac{144}{2652}}$

${\displaystyle P(1,2)=P(A{S}_1{S}_2\bigvee S_{1}A{S}_2)=\frac{1}{52}•\frac{12}{51}+\frac{12}{52}•\frac{1}{51}=\frac{24}{2652}}$

${\displaystyle P(2,0)=P\left(A_1{A}_2\right)=\frac{3}{52}•\frac{2}{51}=\frac{6}{2652}}$

${\displaystyle P(2,1)=P(A{S}_1{A}_2\bigvee A_{1}A{S}_2)=\frac{1}{52}•\frac{3}{51}+\frac{3}{52}•\frac{1}{51}=\frac{6}{2652}}$

$P(2,2)=P\left(\varnothing \right)=0$

Let $A_i,B_i,C_i$ be the events of selecting a sophomore, junior, or senior, respectively, on the i th trial. Let X be the number of sophomores and Y be the number of juniors selected.

Set $P(i,k)=P(X=i,Y=k)$

${\displaystyle P(0,0)=P\left(C_1{C}_2\right)=\frac{3}{8}•\frac{2}{7}=\frac{6}{56}}$

${\displaystyle P(0,1)=P\left(B_1{C}_2\right)+P\left(C_1{B}_2\right)=\frac{3}{8}•\frac{3}{7}+\frac{3}{8}•\frac{3}{7}=\frac{18}{56}}$

${\displaystyle P(0,2)=P\left(B_1{B}_2\right)=\frac{3}{8}•\frac{2}{7}=\frac{6}{56}}$

${\displaystyle P(1,0)=P\left(A_1{C}_2\right)+P\left(C_1{A}_2\right)=\frac{2}{8}•\frac{3}{7}+\frac{3}{8}•\frac{2}{7}=\frac{12}{56}}$

${\displaystyle P(1,1)=P\left(A_1{B}_2\right)+P\left(B_1{A}_2\right)=\frac{2}{8}•\frac{3}{7}+\frac{3}{8}•\frac{2}{7}=\frac{12}{56}}$

${\displaystyle P(2,0)=P\left(A_1{A}_2\right)=\frac{2}{8}•\frac{1}{7}=\frac{2}{56}}$

${\displaystyle P(1,2)=P(2,1)=P(2,2)=0}$

${\displaystyle PX=[30/5624/562/56]PY=[20/5630/566/56]}$

$P(X=i,Y=k)=P(X=i)P(Y=k|X=i)=(1/6\left)P\right(Y=k|X=i)$ .