- #1
johngoogle
- 1
- 0
Suppose
X1 Y1 Z1
0 0 0 (5 times Z1 is 0 for X1=0 and Y1=0)
0 0 0
0 0 0
0 0 0
0 0 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
.
.
.
Which is..
(Same table which is above..)
X1 Y1 Z1 (Count of Zeros) (Count of Ones)
0 0 5 2
0 1 7 8
1 0 0 10
1 1 5 1
X2 Y2 Z2(Count of Zeros) (Count of Ones)
0 0 10 4
0 1 14 16
1 0 0 20
1 1 10 2
Finding P(Z1|X1Y1) AND P(Z2|X2Y2)
HERE it is the same..
example
X1 Y1 Z1 (Probability of Zeros)
0 0 5|7
X2 Y2 Z2 (Probability of Zeros)
0 0 10|14
But what if this is not the case and we have
X2 Y2 Z2(Count of Zeros) (Count of Ones)
0 0 7 5
0 1 12 2
1 0 2 0
1 1 1 5
Than how to know how close is P(Z1|X1Y1) AND P(Z2|X2Y2) using some method..
X1 Y1 Z1
0 0 0 (5 times Z1 is 0 for X1=0 and Y1=0)
0 0 0
0 0 0
0 0 0
0 0 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 0
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
0 1 1
.
.
.
Which is..
(Same table which is above..)
X1 Y1 Z1 (Count of Zeros) (Count of Ones)
0 0 5 2
0 1 7 8
1 0 0 10
1 1 5 1
X2 Y2 Z2(Count of Zeros) (Count of Ones)
0 0 10 4
0 1 14 16
1 0 0 20
1 1 10 2
Finding P(Z1|X1Y1) AND P(Z2|X2Y2)
HERE it is the same..
example
X1 Y1 Z1 (Probability of Zeros)
0 0 5|7
X2 Y2 Z2 (Probability of Zeros)
0 0 10|14
But what if this is not the case and we have
X2 Y2 Z2(Count of Zeros) (Count of Ones)
0 0 7 5
0 1 12 2
1 0 2 0
1 1 1 5
Than how to know how close is P(Z1|X1Y1) AND P(Z2|X2Y2) using some method..