# Probability of the events

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.

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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 $\Bbb P(A\cap B)=1/11$.

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Aha, so I will have to compute it in the following way, P(A intersection B) = 3/11+4/11 - 12/121 = 77/121 - 12/121 = 65/121! –  Takarakaka Jul 2 '12 at 19:09
@Takarakaka: In part (b) are you supposed to find $\Bbb P(A\cup B)$ or $\Bbb P(A\cap B)$? The wording of the question suggests that you want $\Bbb P(A\cap B)$, which is just $12/121$. –  Brian M. Scott Jul 2 '12 at 19:12
Yes, I wanted to solve P( A intesection B), but I hurried up and wrote P( A union B)! What I did there was union, so I will fix it out now. Thanks for your help, @Brian M. Scott –  Takarakaka Jul 2 '12 at 19:13
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 $11/121$.
You’ve misinterpreted the question, I think: the OP is to determine $\Bbb P(A\cap B)$ in each of two separate settings. –  Brian M. Scott Jul 2 '12 at 19:16
@Takarakaka: Under the conditions in (a) the probabilities are not independent; in (b) you’re assuming that they are independent and using that to determine $\Bbb P(A\cap B)$. –  Brian M. Scott Jul 2 '12 at 19:18