联合概率密度函数的问题!P (X=0|Y

网友回答

P ( X = 0 | Y =1 )
=P(X=0,Y=1)/P(Y=1)
=P(X=0,Y=1)/[P(Y=1,X=0)+P(Y=1,X=1)],因为Y=1时,X只有两种可能
=0.21/(0.21+0.19)
=21/40
P ( Y = 0 | X =1 )
=P(Y=0,X=1)/P(X=1)
=P(Y=0,X=1)/[P(X=1,Y=0)+P(X=1,Y=1)],因为X=1时,Y只有两种可能
=0.36/(0.36+0.19)
=0.36/.55
=36/55
======以下答案可供参考======
供参考答案1:
P(X=0|Y=1)=P(X=0,Y=1)/P(Y=1)
=0.21/(0.21+0.19)
=0.525
P(Y=0|X=1)=P(Y=0,X=1)/P(X=1)
=0.36/(0.36+0.19)