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Given are two urns, urn $U_1$ contains $3$ black and $2$ white balls, and urn $U_2$ contains $2$ black and $3$ white balls. A fair coin is flipped to decide which urn we should draw from. We draw $2$ balls from the selected urn with replacement (putting each ball back after we draw it). The question then asks:

$(i)$ What is the probability that the second ball we draw is black, if the first one is also black.

$(ii)$ What is the probability that the second ball is black, if urn $U_1$ is selected and the first ball is black.

$(iii)$ What is the probability that urn $U_1$ was selected, if the first ball is black.

$(iv)$ Given two events $A$ and $B$:

$A:$ $U_1$ is selected and the first ball is black.

$B:$ The second ball is black.

Are $A$ and $B$ independent?


I'm stuck at point $(iii)$. My idea was to define two events for $(i)$:

$A_1:$ The first ball is black.

$A_2: $ The second ball is black.

Since the question does not specify which urn to draw from, then $P(A_1) = P(A_2) = 0.5$ (as our sample space has $10$ balls, $5$ of them are black). Using the conditional probability definition and since $A_1$ and $A_2$ are independant:

$P(A_2 \mid A_1) = \frac{P(A_2 \cap A_1)}{P(A_1)} = \frac{P(A) \cdot P(B)}{P(A_1)} = \frac{0.5 \cdot 0.5}{0.5} = 0.5$.

$(ii)$ Here my approach was to define event $A_3 : $ "Urn $U_1$ is selected" with $P(A_3) = 0.5 $ (since it's a fair coin) and recalculate $P(A_1)$ as our sample space got smaller:

$P(A_1) = \frac{3}{5}$ $\Rightarrow P(A_3 \mid A_1) = 0.5$.

How do I proceed with $(iii)$? If I used the same approach, what would $P(A_3 \cap A_1)$ be? Is my work correct at all? Any help would be much appreciated.

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  • $\begingroup$ (I) $A_1$ and $A_2$ are not independent. They are conditionally independent when given the Urn. You need to consider the urn. $\endgroup$ – Graham Kemp May 5 '20 at 21:28
  • $\begingroup$ @GrahamKemp How can I exactly consider the urn? Do I give two answers for the two possible scenarios (that we draw either from $U_1$ or $U_2$)? How should event $B$ in $(iv)$ be interpreted then? $\endgroup$ – Karla May 5 '20 at 21:43
  • $\begingroup$ Use the Law of Total Probability. $\endgroup$ – Graham Kemp May 5 '20 at 22:43
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If events A and B are conditionally independent over event C (and its complement), then that and the Law of Total Probability states: $$\mathsf P(A, B)=\mathsf P(A\mid C)\mathsf P(B\mid C)\mathsf P(C)+\mathsf P(A\mid C^\complement)\mathsf P(B\mid C^\complement)\mathsf P(C^\complement)$$Use this.


(i) What is the probability that the second ball we draw is black, if the first one is also black.

Using $A_1, A_2$ for "the first/second ball is black" and $U_1,U_2$ for the complementary events of selecting urn 1 or 2 respectively.  Since we are drawing with replacement from the same urn, $A_1,A_2$ are conditionally independent given whichever urn is drawn and have identical conditional probabilities for a given urn.$$\mathsf P(A_1\mid U_1)=\mathsf P(A_2\mid U_1)=3/5\\\mathsf P(A_1\mid U_2)=\mathsf P(A_2\mid U_2)=2/5\\\mathsf P(U_1)=\mathsf P(U_2)=1/2$$

So using Bayes' Rule with the above.

$$\begin{align}\mathsf P(A_2\mid A_1)&=\dfrac{\mathsf P(A_1,A_2)}{\mathsf P(A_1)}\\[2ex]&=\dfrac{\mathsf P(A_1\mid U_1)\mathsf P(A_2\mid U_1)\mathsf P(U_1)+\mathsf P(A_1\mid U_2)\mathsf P(A_2\mid U_2)\mathsf P(U_2)}{\mathsf P(A_1\mid U_1)\mathsf P(U_1)+\mathsf P(A_1\mid U_2)\mathsf P(U_2)}\\[1ex]&=13/25\end{align}$$

(ii) What is the probability that the second ball is black, if urn $U_1$ is selected and the first ball is black.

You seek $\mathsf P(A_2\mid A_1, U_1)$ which you can find by using Bayes' Rule as above. However, here's a hint: "$A_1,A_2$ are conditionally independent give $U_1$".

What is the probability that urn $U_1$ was selected, if the first ball is black.

Again, $\mathsf P(U_1\mid A_1)$ can be found using Bayes' Rule.  However, notice that because there are five balls in each urn, then they are all equally likely to be the first ball drawn, and three of the five black balls are in urn 1.

Given two events $A$ and $B$:

Well, $A=A_1\cap U_1$ and $B=A_2$ so you seek to see if $\mathsf P(A_1,A_2,U_1)$ equals $\mathsf P(A_1,U_1)\mathsf P(A_2)$ or not.

Do that.

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  • $\begingroup$ Thanks a lot for your help. $\endgroup$ – Karla May 6 '20 at 0:58
  • $\begingroup$ Note: $P(A_2)=P(A_2,U_1)+P(A_2,U_2)$. $\endgroup$ – Graham Kemp May 6 '20 at 23:07
  • $\begingroup$ @Karla Yes, indeed. $\endgroup$ – Graham Kemp May 7 '20 at 0:05
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I do not think your answer for (ii) is correct. It is given that the urn $U_1$ is selected. Therefore $P (A_2)=\frac35$.

To approach (iii) let use Bayes theorem: $$ P (U_1|A_1)=\frac {P (U_1\cap A_1)}{P (A_1)}=\frac {\frac12\frac35}{\frac12\frac35+\frac12\frac25}=\frac35. $$

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Let $A_4$ define the event that Urn $2$ is selected. Then, using Bayes’s Theorem, we have

$$P(A_3 | A_1) = \frac{P(A_3)\cdot P(A_1 | A_3)}{P(A_3) \cdot P(A_1 | A_3) + P(A_4) \cdot P(A_1 | A_4)} $$ $$=\frac{0.5 \cdot \frac 35}{0.5\cdot \frac 35 + 0.5 \cdot \frac 25} =\frac 35$$

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