The probability theory problem proves that if P A B P A B is inverse , then there is P B A P B A inv

It just so happened that this question was done on my homework last semester

Agree on the symbol, and the opposite event of a is denoted as a', is the fractional line, e.g. a b represents a of b points.

First, use the inverse formula p(a|b')=p(a)p(b'|a)/p(b')=p(a)/p(b') ×1-p(b|a)) (Because p(b.)'|a)=1-p(b|a)）

With the condition p(a|b)>p(a|b') to get p(a|b)>p(a)/p(b') ×1-p(b|a)) To solve this inequality, put p(a|b) placed on the left, available.

p(a|b) > p(a) (all the other quantities are about to be made, you can calculate it yourself).

Then put p(b|a) and p(b|a'It's all written in an inverse formula.

p(b|a)=p(b)p(a|b)/p(a)，p(b|a')=p(b)p(a'|b)/p(a')

After that, since you want to compare their sizes, then divide them and look at the relationship between the quotient and the size of 1.

p(b|a')/p(b|a)=p(a)/p(a')×p(a'|b)/p(a|b) Reuse p(a'|b)=1-p(a|b) Substitution.

p(b|a')/p(b|a)=p(a)/p(a')×(1-1/p(a|b))

In this case, we need to use the conclusion p(a|) that I have just reached with great difficultyb) > p(a) and replace p(a|) with p(a).b) The right end of the above formula will become larger.

p(b|a')/p(b|a)p(b|a')

Since ab ac=a(b c) is contained in a, then p(ab ac) p(a), on the other hand, there is.

p(ab ∪ac)＝p(ab)+p(ac)-p(ab∩ac)=p(ab)+p(ac)-p(abc) ≥p(ab)+p(ac)-p(bc).

(because p(abc) p(bc)) is obtained by equations (1) and (2).

p(ab)+p(ac)-p(bc)≤ p(a)。PR (Probability) means probability, also known as probability, chance or probability. PR is the basic concept of mathematical probability theory, which is a real number between 0 and 1 and is a measure of the probability of a random event occurring.

The concept of probability is applied in life to represent the probability of a random event occurring, which is an inherent property of the event itself that does not change with people's subjective will.

ab ac bc a(b c) bc when there is only b event, a 0, a(b c) bc 0 i.e. a(b c) bc a when not only b event (1) when there is a event, b c 1, then a(b c) a,a(b c) bc a (2) when there is no a event, a(b c) bc bc 0, because a 0,a(b c) bc a summarized, a(b c) bc a

Proof for arbitrary events a, b, c

Since ab ac=a(b c) is contained in a, then p(ab ac) p(a), 1).

On the other hand, there are.

p(ab ∪ac)＝p(ab)+p(ac)-p(ab∩ac)=p(ab)+p(ac)-p(abc) ≥p(ab)+p(ac)-p(bc).2) (because p(abc) p(bc)).

It can be obtained from equations (1) and (2).

p(ab)+p(ac)-p(bc)≤ p(a),

Analyze on a case-by-case basis, e.g. each is independent of each other.

p(a-b) p(inverse).

Because p(b)=, therefore, p(b is inverse).

According to the probability of stupidity condition.

p(a|b inverse) p (inverse) split mu p(b inverse) , so positive source accompanies p(a-b).

FYI.

p(a-b)=p(a)-p(b)=p(ab inverse), that is, the auspicious deficit domain of region a minus the region of ab intersecting the fibrous domain or the inverse intersection of a and b.

p(a b inverse) = p(ab inverse) p(b inverse) = p(b inverse) = 1-p(b) =

You can ask for a god.

p(ab inverse) =

First of all, you need this:

When a b =

i.e., a,b mutually exclusive): p(a+b)=p(a)+p(b);

The following proves the conclusion given by the question:

Note: When b is included in a, there is:

a=b+a-b)

And b (a-b)=

Therefore there is: p(a) = p(b) + p(a-b).

So there is the following conclusion: [p(a-b)=p(a)-p(b)].

And when there is no condition where b is included in a: then it is due to: a-b=a-ab

AB is contained in A. Therefore:

Thus there is p(a-b)=p(a-ab).

p(a)p(ab)

Difference: p(a-b) = p(a)-p(ab) applies in all cases p(a-b) = p(a)-p(b).

This is true only if condition B is included in A.

Connection: Actually, the former is just a variation of the latter.

Certificate: p(a)-p(ab)=0

By ab necessarily contains a

The above formula is: p(a-ab)=p(ac)=0

Write non-b as c).

Because an event with a probability of 0 is not necessarily an impossible event (such as a point in a uniform distribution), AC

Not necessarily. Impossible event.

So A is not necessarily included in B

All are valid, but conditions are required:

1、p(a-b)=p(a)-p(b) ：

In probability theory, there is equality of probabilities before events are equal.

Due to the monotonicity of probability, p(a-b)=p(a)-p(b) is true only if the condition "b is contained in a" is true.

For any two events A and B, B is not necessarily included in A, and AB must be included in A, so a-b = a-ab, so: p(a-b) = p(a)-p(ab).

2、p(a+b)=p(a)+p(b) ：

A sufficient and necessary condition for ab mutex is that p(a+b) = p(a)+p(b) and that the intersection of p(a) and p(b) is not an empty set.

Let the number of occurrences of random event A in n repeated trials be na, if the frequency na n swings steadily around a certain value p when the number of trials n is large, and the amplitude of the swing becomes smaller and smaller with the increase of the number of trials n, then the number p is called the probability of random event a, which is denoted as p(a)=p.

Evidence: Because p(a|b) >p(a), p(a) cannot be 0, otherwise a is a zero generalized reed rate event, p(a|b)=p(a)=0 So there is p(a|b)/pa>1 p(b|a)=p(ab)/p（a）=p(b)p(a|b) pa>p(b) If you can understand it, I just want to talk about it briefly, and I don't need to dwell on this issue to guess that they are not independent events (otherwise.

But it's okay to be cautious.

p(ab)=p（b\a）p(a)

p(b a) denotes the probability that b will be born under the condition that a occurs. Because b must occur, it also occurs under the condition that a occurs, that is, p(b a)=1

So p(ab) = p(a).