Probability - Independence Question

jmcgraw
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I'm trying to get the idea of independent events well grounded in my mind, but I'm having some difficulty.

First of all, what would a venn diagram for independent events look like? I know you have the two events intersecting, and the intersection is equal to the product of the probabilities of the two events... But all I can picture is a normal diagram of two events intersecting, which by eyeballing looks just like the diagram of any 2 events (depndent or independent). Is there something unique to the diagram of independent events?

Mathematically, I just need to know that P(A \cap B) = P(A)P(B) . But I can't form a graphical intuition about this, which I think would help me a lot.

Incidentally, I can't help but keep picturing the venn diagram of mutually exclusive events as a representation of independent events... But I know that is dead wrong, since mutually exclusive events are dependent.

Can anyone help me clarify this in my mind? Thanks!
 
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If B is independent of A, it means that the intersection of B with A is the SAME fraction of A's total area as the intersection of B with not-A is of not-A's area.
 
If B is independent of A, it means that the intersection of B with A is the SAME fraction of A's total area as the intersection of B with not-A is of not-A's area.

I think I just got more confused. :smile:

Isn't the intersection of B with not-A the same as not-A? giving the fraction to be simply 1?

Obviously I must be missing something, big time.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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