Answer to Question #173473 in Statistics and Probability for ANJU JAYACHANDRAN

Question

5(b) For the given bivariate probability distribution of X and Y : (5) 

 

32

( , )

2

x y

P X x Y y

+

= = = for x = 3,2,1,0 and y = .1,0

 Find: 

 (i) P(X ≤ ,1 Y = )1

 (ii) P(X ≤ )1

 (iii) P(Y > )0 and 

 (iv) P(Y = |1 X = )3 

Answer 1

(i) $P(X \le ,1 Y =1)=P_{XY}(1,1)=0$

(ii) $P(X\le 1)=P_{XY}(1,2)+P_{XY}(1,4)+P_{XY}(1,5)$

$=\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{24}=\dfrac{2+1+1}{24}=\dfrac{4}{24}=\dfrac{1}{6}$

(iii) $P(Y>0)=P_{XY}(2,1)+P_{XY}(2,2)+P_{XY}(2,3)+P_{XY}(4,1)+P_{XY}(4,2)+P_{XY}(4,3)+P_{XY}(5,1)+P_{XY}(5,2)+P_{XY}(5,3)$

$=\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{12}+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{8}+\dfrac{1}{12}$

$=\dfrac{2+4+6+1+2+3+1+3+2}{24}=\dfrac{24}{24}=1$

(iv) $P(Y=1|X=3)=\dfrac{P(X=3,Y=1)}{P(X=3)}$

$=\dfrac{P_{XY}(3,1)}{P_X(3)}$

$=\dfrac{0}{P_{X}(0)}=0$