Probabilty theory - question about two independent events

Nov 2012
164
2
israel
If two events \(\displaystyle A\) and \(\displaystyle B\) are mutually exclusive (\(\displaystyle A\cap B=\phi\) ), then \(\displaystyle A\) and \(\displaystyle B\) are dependent.
It implies that if \(\displaystyle A\) and \(\displaystyle B\) are independent, they are not mutually exclusive.

Is it true that if \(\displaystyle A\) and \(\displaystyle B\) are independent, then either \(\displaystyle A\subseteq B\) or \(\displaystyle B\subseteq A\)?
 
Last edited:

chiro

MHF Helper
Sep 2012
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Australia
Hey Stormey.

We have P(A OR B) = P(A) + P(B) - P(A and B) = P(A) + P(B) since A and B = null event and P(Null event) = 0 (since P(Not Null) = 1). We also have independence to be defined as P(A|B) = P(A) and P(A and B) = P(A)P(B).

Recall though that since A and B is the null event, then P(A and B) = 0 regardless of what A and B are. This violates the assumption of independence since P(A|B) = P(A) will be strictly positive and depend on the actual event (instead of being zero).

The other intuitive thing is that independence means that knowing one thing about another variable means no advantage when it comes to knowing something about the variable of interest. Since we know that both variables are disjoint, this implies that knowing something about the other directly tells us about the variable of interest since they are disjoint.

There are a few ways to look at this but hopefully the above has given you further insight.
 
Nov 2012
164
2
israel
Hi chiro, thanks for your help.
I'm taking my first steps in the theory of probability, and it is very important for me, at least at first, that the mathematical definitions and theorems will capture my intuition about what I already know about probability from day to day experience (and that is not much, I must admit).

So...
This is how I picture it in my mind.
Suppose that A and B are some disjoint events in the sample space:



then if A heppened, and we then want to calculate the probability that B will also happen, we have to consider, in our calculation, the fact that some outcomes (in particular - those that in event A) have already happened, thus reducing the sample space.

for example:
Suppose we have a box with 9 balls: 4 blue balls, and 5 red balls.
Two balls are randomly picked from the box.
What would be the probability to draw a blue ball and then a red ball (with NO returning them back)?
So the sample space is \(\displaystyle \Omega =\left \{ drawing\hspace{5}any\hspace{5}ball\hspace{5}from \hspace{5} the \hspace{5} box \right \}\), and then \(\displaystyle |\Omega |=9\)
and in this case, \(\displaystyle A=\left \{ drawing\hspace{5}a\hspace{5}blue\hspace{5}ball \right \}\), \(\displaystyle B=\left\{ drawing\hspace{5}a\hspace{5}red\hspace{5}ball \right \}\)
so if \(\displaystyle A\) happened (we drew a blue ball) it'll make the sample space shrink (we're not returning the blue ball back):
\(\displaystyle \mathbb{P}(A)=\frac{4}{9}\)
and then:
\(\displaystyle \mathbb{P}(B)=\frac{5}{8}\)
So
\(\displaystyle \mathbb{P}(A\cap B)=\frac{4}{9}\cdot\frac{5}{8}\)
In the above case, A and B are mutually exclusive (you can't draw blue and red both at the same time, with only one pull), therefore dependent. so far so good.

but, and this is where it gets confusing for me; what happens if there are two independent events?



can you please give me an example, based on the scenario above (with the box and the balls), of two independent events, where the occurrence of the first one does not effect the probability of the other?
I might say "well, the same as above, only this time we DO return the ball after the first pick; this way, the occurrence of the first one does not effect the probability of the other..." then we get that A and B are indeed independent, BUT(!) A and B are still mutually exclusive! and that's impossible, according to what I wrote in my first post in this thread.
 
Last edited:

Plato

MHF Helper
Aug 2006
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Suppose we have a box with 9 balls: 4 blue balls, and 5 red balls.
Two balls are randomly picked from the box.
What would be the probability to draw a blue ball and then a red ball (with NO returning them back)?
So the sample space is \(\displaystyle \Omega =\left \{ drawing\hspace{5}any\hspace{5}ball\hspace{5}from \hspace{5} the \hspace{5} box \right \}\), and then \(\displaystyle |\Omega |=9\)
and in this case, \(\displaystyle A=\left \{ drawing\hspace{5}a\hspace{5}blue\hspace{5}ball \right \}\), \(\displaystyle B=\left\{ drawing\hspace{5}a\hspace{5}red\hspace{5}ball \right \}\)
so if \(\displaystyle A\) happened (we drew a blue ball) it'll make the sample space shrink (we're not returning the blue ball back):
\(\displaystyle \mathbb{P}(A)=\frac{4}{9}\)
and then:
\(\displaystyle \mathbb{P}(B)=\frac{5}{8}\)
So
\(\displaystyle \mathbb{P}(A\cap B)=\frac{4}{9}\cdot\frac{5}{8}\)
In the above case, A and B are mutually exclusive (you can't draw blue and red both at the same time, with only one pull), therefore dependent. so far so good.
but, and this is where it gets confusing for me; what happens if there are two independent events?
Let's be clear. Event \(\displaystyle A\) is the first ball is blue and the event \(\displaystyle B\) is the second ball is red.
Then you are correct \(\displaystyle \mathcal{P}(A\cap B)=\frac{4}{9}\cdot\frac{5}{8}\)

But you cannot say \(\displaystyle \mathcal{P}(B)=\frac{5}{8}\), you can say \(\displaystyle \mathcal{P}(B|A)=\frac{5}{8}\).

You see that \(\displaystyle \mathcal{P}(B)=\mathcal{P}(B\cap A)+ \mathcal{P}(B\cap A^c)=\frac{5}{9}\)

NOW you see that \(\displaystyle \mathcal{P}(B|A)\ne\mathcal{P}(B)\) therefore those two events are not independent.

If you would return the ball back to the box after each draw then the two events are independent.
 
Nov 2012
164
2
israel
Hi Plato, thanks for the comment.

You see that \(\displaystyle \mathcal{P}(B)=\mathcal{P}(B\cap A)+ \mathcal{P}(B\cap A^c)=\frac{5}{9}\)
Why is \(\displaystyle \mathcal{P}(B)=\mathcal{P}(B\cap A)+ \mathcal{P}(B\cap A^c)\)?

If you would return the ball back to the box after each draw then the two events are independent.
So in what way are these two events NOT mutualy exclusive?
I mean, why are A (which is "drawing first ball blue") and B (which is "drawing second ball blue") not mutually exclusive?
what exactly is mutual between them?
 

Plato

MHF Helper
Aug 2006
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8,653
Why is? So in what way are these two events NOT mutually exclusive?
I mean, why are A (which is "drawing first ball blue") and B (which is "drawing second ball blue") not mutually exclusive?
what exactly is mutual between them?
They are NOT mutualy exclusive events!

Is it possible that the first ball is blue(A) and the second ball is red(B) at the same time?

The criteria for two events to be mutually exclusive is that it is impossible for both to occur.

The two events \(\displaystyle B_1R_2~\&~R_1R_2\) are mutually exclusive events; drawing two balls the first cannot be both blue(\(\displaystyle B_1\)) and at the same time be red(\(\displaystyle R_1\)).
 
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