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Let $ A $ and $ B $ be events such that $ P(A).P(B) > 0, $ then $ A $ and $ B $ are independent iff $ P(A|B) = P(A), $ or equivalently $ P(A \cap B) = P(A).P(B). $

$ \textbf{Question:} $ An urn contains $ m $ red balls and $ n $ blue balls and at each stage we pick a ball from the urn without replacement. Let $ A $ be the event that the first ball chosen is red and $ B $ be the event that the second ball chosen is red. Are $ A $ and $ B $ independent?

I compute $ \displaystyle P(A \cap B) = \frac{m}{m + n}. \frac{m - 1}{m + n - 1} $ and $ \displaystyle P(A).P(B) = \frac{m}{m + n}. \frac{m}{m + n - 1} $ and so $ A $ and $ B $ are not independent.

Am I doing it correctly? I am confused about the values of $ P(B) $ since it depends on whether $ A $ happens or not. If $ A $ happens, then $ \displaystyle P(B) = \frac{m - 1}{m + n - 1}, $ otherwise $ \displaystyle P(B) = \frac{m}{m + n - 1}. $

Also, is it true that $ \displaystyle P(A \cap B) = \frac{m}{m + n}. \frac{m - 1}{m + n - 1} $ in this case? I put that computation since it makes me think of an analogue where I flip a coin twice and get $ \displaystyle P({head \; first \; turn} \cap {tail \; second \; turn}) = \frac{1}{2}. \frac{1}{2} = \frac{1}{4}. $ However, in the ball and urn case $ A $ and $ B $ are not independent.

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    $\begingroup$ You're indeed wrong on the computation of $P(B)$. To be able to compute it safely, you can compute $P(A \cap B) + P(($not $A) \cap B)$ $\endgroup$ – WNG Oct 1 '16 at 19:15
  • $\begingroup$ Follow-up question: math.stackexchange.com/questions/1949653/… $\endgroup$ – yurnero Oct 1 '16 at 20:00
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As you point out: $B$ and $A$ are not independent because conditioning on whether $A$ happens changes the conditional probability of $B$: \begin{align*} \Pr(B)&=\Pr(B\cap A)+\Pr(B\cap\neg A)\\ &=\Pr(B\mid A)\Pr(A)+\Pr(B\mid\neg A)\Pr(\neg A)\\ &=\frac{m-1}{m+n-1}\frac{m}{m+n}+\frac{m}{m+n-1}\frac{n}{m+n}\\ &=\frac{m}{m+n}. \end{align*} Your confusion seems to stem from your casual language not distinguishing among $\Pr(B)$, $\Pr(B\mid A)$ and $\Pr(B\mid\neg A).$

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Solution: I think it will be wise to calculate the conditional probability $P(B|A)$. In other words, we are finding the probability of the Event $B$ happening given that Event $A$ has occurred. Now if $P(B|A)=P(B)$, then we can conclude that the two Events are independent. Now, $$P(B|A)=\frac{m-1}{m+n-1}$$ On the other hand, $$P(B)=P(\text{first ball is red and the second ball is red})+P(\text{first ball is blue and the second ball is red})$$$$=\frac{m}{m+n}*\frac{m-1}{m+n-1}+\frac{n}{m+n}*\frac{m}{m+n-1}$$ $$=\frac{m^2-m+mn}{(m+n)(m+n-1)}\neq P(B|A)$$ Thus, we can conclude that since $P(B|A)\neq P(B)$, the two Events are indeed independent.

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  • $\begingroup$ Should $ \displaystyle P(B|A) = \frac{m - 1}{m + n - 1} $ only? Where does the term $ \displaystyle \frac{m}{m + n} $ come from? $\endgroup$ – user298251 Oct 1 '16 at 19:31
  • $\begingroup$ Oops.. I was in a rush. Will edit. $\endgroup$ – model_checker Oct 1 '16 at 19:34

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