(in)dependent events?

**courteous** · August 25th 2010, 11:53 AM

We have 4 cards on a table: $\spadesuit K$ , $\spadesuit Q$ , $\heartsuit K$ and $\heartsuit Q$ .
We blindly choose one card.
Define events as:
$A \equiv \spadesuit$ chosen
$B \equiv Q$ chosen
$C \equiv \heartsuit K \cup \spadesuit Q$ chosen

Are following events independent?

$A \cap B$
$A \cap C$
$B \cap C$
$A \cap B \cap C$

Yes.
Yes: $P(A \cap C)=P(A)*P(C|A)=\frac{1}{2}\frac{1}{2}==P(A)*P(C)=\ frac{1}{2}\frac{1}{2}$
Yes: $P(B \cap C)=P(B)*P(C|B)=\frac{1}{2}\frac{1}{2}==P(B)*P(C)=\ frac{1}{2}\frac{1}{2}$ [Solutions say "No."]
No: $P(A \cap B \cap C)=P(A)*P(B|A)*P(C|A\cap B)=\frac{1}{2}\frac{1}{2}1\neq P(A)*P(B)*P(C)=\frac{1}{2}\frac{1}{2}\frac{1}{2}$

Please, also tell me whether points 2. and 4. are correct (for the right reasons)?

Thank you!

**awkward** · August 25th 2010, 01:45 PM

Hi Courteous,

I think you are correct on all 4 parts.

However, I think you have sometimes done things the hard way. Rather than computing $P(A \cap C)$ as $P(A) * P(C | A)$ , for example, it's easier to observe that $A \cap C$ is the event of drawing the queen of spades, 1 card out of 4, hence its probability is 1/4.

**Plato** · August 25th 2010, 03:01 PM

I think some of the confusion in the OP comes from notation.
Express the events in set notation: $A = \left\{ {\spadesuit K,\spadesuit Q} \right\},\,B = \left\{ {\spadesuit Q,\heartsuit Q} \right\}\;\& \,C = \left\{ {\heartsuit K,\spadesuit Q} \right\}$
It becomes clear that $P(A)=P(B)=P(C)=0.5$ .

Now the question as to if events $A~\&~C$ are independent comes down to asking if $P\left( {A \cap C} \right)\mathop = \limits^? P(A)P(C)$ .
But $A\cap C=\left\{ {\spadesuit Q} \right\}$ which has probability $0.25$ .

The question about $A,~B,~\&~C$ being independent is actually vague.
It is clear that $P\left( {A \cap B \cap C} \right) \ne P(A)P(B)P(C)$ . Is that what is meant?

I also do not understand the bit in red.
If the solution set says that, it is wrong.

**courteous** · August 25th 2010, 11:01 PM

Originally Posted by Plato

The question about $A,~B,~\&~C$ being independent is actually vague.
It is clear that $P\left( {A \cap B \cap C} \right) \ne P(A)P(B)P(C)$ . Is that what is meant?

Instructions in OP are as given, so I know even less.

I would like to make another "are (in)dependent?" problem:

We have 8 cards: $A, K, Q, J, 10, 9, 8, 7$ .
We pick one.
Events:
$A=\left\{A,K,Q,J\right\}$
$B=\left\{A,9,8,7\right\}$
$C=\left\{A,K,Q,10\right\}$
Are events $A$ , $B$ and $C$ independent?

$P(A \cap B \cap C)=\frac{1}{8}==P(A)*P(B)*P(C)=\frac{1}{2}\frac{1} {2}\frac{1}{2}$
Solutions say "No, they're not independent (even though $P(A \cap B \cap C)=P(A) P(B) P(C)$ )."
Seems reasonable to me also (looking at events' definitions), but how to argue that, indeed they're not independent?

Does independency require that all pairs are also independent (that, say, $P(C|A)=P(C)$ , which is obviously not true)?

**Plato** · August 26th 2010, 06:00 AM

Originally Posted by courteous

$P(A \cap B \cap )=\frac{1}{8}==P(A)*P(B)*P(C)=\frac{1}{2}\frac{1}{ 2}\frac{1}{2}$
Solutions say "No, they're not independent (even though $P(A \cap B \cap C)=P(A) P(B) P(C)$ )."
Does independency require that all pairs are also independent (that, say, $P(C|A)=P(C)$ , which is obviously not true)?

Yes, you have answered your own question.
Each subcollection has to be independent. So each pair must be independent.

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