# statistics, probability, event independency

As I'm trying to figure out what this is all about I would like somebody to take a look at my effort here and possibly correct me if I'm wrong.

\begin{align} & P(A) = 1/2 \\ & P(B) = 1/3 \end{align}

Calculate: $P( A ∪ B )$

So:
I)
$P( A ∪ B ) = P(A) + P(B) - P( A ∩ B )$

$P( A ∩ B ) = 1/2 * 1/3 = 1/6$

$1/2 + 1/3 - 1/6 = 2/3$

II)$P( A ∩ B ) = 1/7$ - are $A$ and $B$ independent?

Answer:
$1/7$ is not equal $1/3 * 1/2$
$1/7$ is not equal $1/6$

$A$ and $B$ are not independent.

Am I even doing this correctly? My apologies but I'm completely lost.

• You can not calculate $P(A\cup B)$ without knowing about the possible dependence. Suppose you are throwing a fair die, and that $A$ is the event "you throw an even number" and $B$ is the event "you throw a $2$ or a $4$". Then $P(A\cup B)=\frac 12$. On the other hand, suppose $B'$ is the event "you throw a $1$ or a $3$." then $P(A\cup B')=\frac 56$.
– lulu
Sep 5, 2015 at 18:04

## 1 Answer

In I) you did correct in using the formula $P(A \cup B)= P(A)+P(B)-P(A \cap B)$ but the problem is not fully solvable without knowing $P(A \cap B)$. You assumed that $P(A \cap B)=P(A)P(B)$ but that holds if and only if they are independent.

II) is correct