Question: For the events of $A$ and $B$, probabilites are $\Bbb P(A) = 3/11$ and $\Bbb P(B) = 4/11$. Define the $\Bbb P(A \cap B )$ if:

a) $\Bbb P(A \cup B) = 6/11$.

b) events are indenpendent

I have done a task, and it's following

$$\Bbb P(A \cup B) = \Bbb P(A) + \Bbb P(B) - \Bbb P(A \cap B ) = 1/11\;.$$

However I don't know how to solve b task. I need your help. 

Thanks in advance.

Answer: 
To say that $A$ and $B$ are independent is to say that $\Bbb P(A\cap B)=\Bbb P(A)\Bbb P(B)$.

By the way, it’s not correct to write

> 
> $\Bbb P(A\cup B)=\Bbb P(A)+\Bbb P(B)-\Bbb P(A\cap B)=1/11$;
> 
> 
> 

what you mean is that $\Bbb P(A\cup B)=\Bbb P(A)+\Bbb P(B)-\Bbb P(A\cap B)$, so $\Bbb P(A\cap B)=1/11$.
Independence means $P(A \cap B)=P(A) P(B)$ but $P(A) P(B)=3 \times 4/121 \approx 0.0991$ and you found $P(A \cap B)=1/11 \approx 0.09091$. Close but not equal, so not independent. Note the difference between $12/121$ vs $11/121$.