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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 586

In order to consistently define the values of p(a) and p(b) why p(a&b)

≤ p(a) and p(b)?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,914

hi 666 bro

I'm not quite sure what you're asking so here's some general help with this sort of probability. I'm re-using an old diagram to save making a fresh one as it will work for this. Let's say that a=bread and b=milk.

There are 100 houses in an area. Some have bread delivered; some have milk; some have both; and some have neither. If you pick a house at random what's the probability of …..

p(a) ie. probability of picking a house that has bread delivered = 60/100

p(b) = ie. the probablilty of picking a house that has milk delivered = 50/100

p(a OR b) = ie. the probability of picking a house that has either bread or milk or both = 80/100

p(a AND b) = ie. the probability of picking a house that has both delivered = 30/100

You can see that p(a OR b) = p(a) + p(b) - p(a AND b) [80/100 = 60/100 + 50/100 - 30/100]

This will always be true because when you add p(a) to p(b) you get all of a, all of b but you count (a AND b) twice.

As (a AND b) is contained in both a and b it will never be greater than their sum. It could be equal if, say, the region of b that doesn't overlap a is empty.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 586

I'm asking what are the general conditions for defining the existing value of p(a&b)?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,914

I'm still unclear.

a must be an event for which a probability can be estimated or determined.

b similarly.

In that case a & b exists as a measurable event. It means both a and b happen.

Look here: https://www.mathsisfun.com/data/probabi … types.html

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,379

The logical definition of A&B is that both of them are true.

Thus from a logical point of view, !A and A cannot be true simultaneously. This is by definition of logic.

also !A+A = ALL

B = (!A U A)&B = ( !A&B )U (A&B)

The first equal sign is self-explanatory, here is the rational for the second:

If a case belongs to !A&B, it must belong to (!A U A)&B since !A U A contains !A

same for A&B

Therefore the left contains the right

you can then prove the right contains the left from a similar logic.

Therefore

B contains A&B.

*Last edited by George,Y (2020-02-01 01:36:26)*

**X'(y-Xβ)=0**

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