Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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

share|cite|improve this answer
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 $11/121$.

share|cite|improve this answer
I see, so this probabilites are not independent. – Takarakaka Jul 2 '12 at 19:16
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
@BrianM.Scott I didn't misprint it, I wrote it correctly, but I now figure out what you are trying to say me! Thank you very much, both of you. – Takarakaka Jul 2 '12 at 19:21
@BrianM.Scott As I reread the question your interpretation seems right. The problem is to calculate P(A interesection B) under 2 different scenarios. In the first case it would be 1/11 and in the second 12/121. My solution is to the problem are A and B independent if P(A)=3/11, P(B)=4/11 and P(AUB)=6/11. – Michael Chernick Jul 2 '12 at 19:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.