Strange as it may sound, since P(B|A) = P(B) the events are independent.

>Correct me if I am wrong?

But stranger things may happen. Consider

A₊ = {1, 2, 3, 5, 7} and
A_- = {3, 5, 7}

and B as before.

Then

P(B|A_-) = 1/3 > 1/4 = P(B), which is sensible I think, but still wonderful. The element "1" has nothing to do with B. Why should its inclusion or exclusion in A affect, even increase, the probability of B.

Worse,

P(B|A₊) = 1/5 < 1/4 = P(B).

And tell me what "2" has to do with B?

The problem is that independence of two events is often confused with them having empty intersection. Your example shows that this is not correct.

Furthermore, having an empty intersection does not imply independence. An event and its complementary are clearly dependent.

Thus, just accept the definition of independence:

P(B|A) = P(B) or

P(A intersection B) = P(A)P(B).

I do not know whether there is anything special about the union of independent events. An elementary event is a 1-element set that contains a single outcome of an experiment. Except for an empty event, all other events are unions of elementary events - possible outcomes. Union of two events is just another event whether the two are independent ot not.

These are the same notational assumptions used in Boolean algebra, where the name of a sentence is used as a variable standing for its truth value, 0 or 1. The notation has merely been extended to use the name of a set (which could have been defined by a sentence anyway) to stand for its probability, 0 <= p <= 1. In fact, it is an assumption of Boolian algebra that the truth values of the variables are independent, but in the context of logic, this assumption is vacuously true, and so passes unnoticed. The assumption of independence becomes visible when you pass from logic to probability.

This perhaps does not answer the original question, but I hope it is at least somewhat useful to mention an interesting feature of independent sets.

--Stuart Anderson

for A-
p(A-) = 3/8
p(B) = 1/4
A intersection B is {7}
p(A intersection B) = 1/8 which is not equal to p(A-).p(B) = 3/32
by which it says A- and B are not independent
will that not same hold good for A and B( that is non independence condition)??

I do not quite understand your question. Just do with A what you did with A+:

P(A) = 1/2, P(B) = 1/4, P(A∩B) = 1/8 = P(A)·P(B)

meaning A and B are independent.

For the example below with prior knowledge of events being independent:
event P ={H,T}-coin
event Q = {1,2,3,4,5,6}-dice
P(p∩q) = P(p)·P(q)= 1/12
Now if Q changes = {1,2,3,4,5,6,7,8}
P(p∩q) = P(p)·P(q)= 1/16
We can do this because we know they are independent.

One more example
Sample Space ={1,2,3,4,5,6,7,8}
Event A={1,3,5,7}, Event B = {5,6,7,8}, P(A)= 1/2,P(B)=1/2,
A intersection B = {5,7},P(A intersection B) =1/4, which is acciddently equal to P(A)*P(B)

My question is
For independent events, P(A∩B) = P(A)·P(B)
but when P(A)·P(B)= P(A∩B) they can independent or non independent.
Please help me to get rid of this question.

>Now as A changed to A+
>or A-, P(A¡ÉB) = P(A)¡¤P(B) does not holds good. But it
>satisfies for A and B, so how can A and B be independent
>when A+ and B are not independent and also A- and B are not
>independent?

However implausible this may look, this is simply true by definition. The definition in this case is not fostered by the pedestrian intuition.

>For the example below with prior knowledge of events being
>independent:
>event P ={H,T}-coin
>event Q = {1,2,3,4,5,6}-dice
>P(p∩q) = P(p)·P(q)= 1/12
>Now if Q changes = {1,2,3,4,5,6,7,8}
>P(p∩q) = P(p)·P(q)= 1/16
>We can do this because we know they are independent.

I would distinguish here "independent events" that come from the same sample space and "independent experiments" that are accommpanied by possibly different sample spaces. The nature of the change in the second component of your example is unrelated to the change of the former sort (A vs A+).

>
>One more example
>Sample Space ={1,2,3,4,5,6,7,8}
>Event A={1,3,5,7}, Event B = {5,6,7,8}, P(A)= 1/2,P(B)=1/2,
>A intersection B = {5,7},P(A intersection B) =1/4, which is
>acciddently equal to P(A)·P(B)

Accidently? Is not everything?

>My question is
>For independent events, P(A∩B) = P(A)·P(B)
>but when P(A)·P(B)= P(A∩B) they can independent or non
>independent.

No, the events A and B are independent iff P(A∩B) = P(A)·P(B). "Only when" means exactly "iff".

>Please help me to get rid of this question