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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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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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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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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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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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