Suppose(adsbygoogle = window.adsbygoogle || []).push({});

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..

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Statistical Comparision between Conditional Probability

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**