Probability, dependent Events I am trying to understand some of the tasks and would be happy if someone could check my solution and possibly point out any mistakes.
In advance, I am not asking for a solution, because I would like to work it out by myself.
The purpose of this exercise is to check whether events A and B are independent of each other.
Three dices are thrown.
The events are defined as follows: A: All dice have the same number of dots.
B: The total sum of the dices is less than 5.
Therefore is
$A = \left \{ \left ( 1,1,1 \right ), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6) \right \}$
$B = \left \{ \left ( 1,1,1 \right ), (1,1,2), (1,2,1), (2,1,1) \right \}$
Some intermediate results:
$P\left ( A \right ) = \frac{6}{216} = \frac{3}{108}$
$P\left ( not A \right ) = 1 - P(A) = 1 - \frac{3}{108} = \frac{35}{36}$
$P(B) = \frac{4}{216} = \frac{1}{54}$
$P(A\cap B) = \frac{1}{216}$
$P(notA\cap B) = \frac{3}{216}$
A and B are independent if
$P(B|notA) = P(B|A)$
$
P(B|A) = \frac{P(A\cap B)}{P(A)} = \frac{\frac{1}{216}}{\frac{6}{216}} = \frac{1}{6}
$
$
P(B|notA) = \frac{P(notA\cap B)}{P(notA)} = \frac{\frac{3}{216}}{\frac{35}{36}} = \frac{1}{70}
$
Thus, the events A and B are dependent.
Thank you in advance for your time.
A: Yes, it seems correct to me.
There can be some more straightforward ways to verify that, you could:

*

*verify if $P(A \cap B) = P(A)P(B)$ that is the definition of independence and perhaps the best way

*verify what you did, or, more generally, that $P(B|A) = P(A)$ or $P(A|B)=P(A)$ (you can also condition on $\bar{A}$ or $\bar{B}$) as independence means that the other event gives you no other information besides what you already have

In general, if two events are independent, then, also all the possible combinations of negations of the events are independent, therefore you can verify equivalently with just one of cases and that is enough.
To summarize, the following statements are equivalent (I indicate with $ A \perp \!\!\! \perp B$ the independence between A and B):

*

*$A \perp \!\!\! \perp B$

*$\bar{A} \perp \!\!\! \perp B$

*$A \perp \!\!\! \perp \bar{B}$

*$\bar{A} \perp \!\!\! \perp \bar{B}$

*$P(A|B)=P(A)$

*$P(B|A)=P(B)$
So, based on the data that you have, you can pick the one that better suits them and use it.
