# Are events $A$ and $B$ independent?

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.

• 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 $A) \cap B)$ – WNG Oct 1 '16 at 19:15
• Follow-up question: math.stackexchange.com/questions/1949653/… – yurnero Oct 1 '16 at 20:00

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 $\Pr(B\mid\neg A).$

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.

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