# (in)dependent events?

• Aug 25th 2010, 11:53 AM
courteous
(in)dependent events?
Quote:

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

Are following events independent?

1. $\displaystyle A \cap B$
2. $\displaystyle A \cap C$
3. $\displaystyle B \cap C$
4. $\displaystyle A \cap B \cap C$

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

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

Thank you!
• Aug 25th 2010, 01:45 PM
awkward
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 $\displaystyle P(A \cap C)$ as $\displaystyle P(A) * P(C | A)$, for example, it's easier to observe that $\displaystyle A \cap C$ is the event of drawing the queen of spades, 1 card out of 4, hence its probability is 1/4.
• Aug 25th 2010, 03:01 PM
Plato
I think some of the confusion in the OP comes from notation.
Express the events in set notation: $\displaystyle A = \left\{ {\spadesuit K,\spadesuit Q} \right\},\,B = \left\{ {\spadesuit Q,\heartsuit Q} \right\}\;\& \,C = \left\{ {\heartsuit K,\spadesuit Q} \right\}$
It becomes clear that $\displaystyle P(A)=P(B)=P(C)=0.5$.

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

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

I also do not understand the bit in red.
If the solution set says that, it is wrong.
• Aug 25th 2010, 11:01 PM
courteous
Quote:

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

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

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

We have 8 cards: $\displaystyle A, K, Q, J, 10, 9, 8, 7$.
We pick one.
Events:
$\displaystyle A=\left\{A,K,Q,J\right\}$
$\displaystyle B=\left\{A,9,8,7\right\}$
$\displaystyle C=\left\{A,K,Q,10\right\}$
Are events $\displaystyle A$, $\displaystyle B$ and $\displaystyle C$ independent?
$\displaystyle P(A \cap B \cap C)=\frac{1}{8}==P(A)*P(B)*P(C)=\frac{1}{2}\frac{1} {2}\frac{1}{2}$
Solutions say "No, they're not independent (even though $\displaystyle P(A \cap B \cap C)=P(A) P(B) P(C)$)."
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, $\displaystyle P(C|A)=P(C)$, which is obviously not true)?
• Aug 26th 2010, 06:00 AM
Plato
Quote:

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