Answer to Question #202258 in Statistics and Probability for RAGHAV SOOD

Question

For the given bivariate probability distribution of X and Y :

 

P(X=x , Y=y)=(x²+y)/32 for x=0,1,2,3 and y=0,1.

 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) $P(X\leq1,Y=1)=P(X=0,Y=1)+P(X=1,Y=1)=\frac{1}{32}+\frac{2}{32}=\frac{3}{32}$;

(ii) $P(X\leq1)=P(X\leq1,Y=1)+P(X\leq1,Y=0)=\frac{3}{32}+P(X=1,Y=0)+P(X=0,Y=0)=\frac{3}{32}+\frac{1}{32}=\frac{1}{8}$ (iii) $P(Y>0)=P(Y=1)=\sum_{x=0}^3P(X=x,Y=1)=\frac{1}{32}+\frac{1}{16}+\frac{5}{32}+\frac{10}{32}=\frac{18}{32}=\frac{9}{16}$

(iv) $P(Y=1|X=3)=\frac{P((Y=1)\cap(X=3))}{P(X=3)}=\frac{P(Y=1,X=3)}{P(X=3,Y=0)+P(X=3,Y=1)}=\frac{\frac{10}{32}}{\frac{9}{32}+\frac{10}{32}}=\frac{10}{19}$