2
$\begingroup$

Urn $1$ contains $2$ black balls and $5$ white balls. Urn $2$ contains $3$ black balls and $2$ white balls. One of the urns is chosen at random and a ball is drawn. The ball is then put in the other urn. From the urn in which the ball was deposited a second ball is drawn. What is the probability that both balls are white?

The solution is $23/70$, but I get $39/112$.
I define the following events: $$\begin{align*} U_i &= \text{the $i$-th urn is selected}\\ W_i &= \text{a white ball is drawn at the $i$-th extraction}\\ \end{align*}$$ with $i=1,2$.

Now, let $p$ be the sought probability. Then $$p = P(W_1 \cap W_2) = P(W_1)P(W_2 \mid W_1).$$ Since $U_1,U_2$ form a partition I can write $$P(W_1) = P(W_1 \mid U_1)P(U_1) + P(W_1 \mid U_2)P(U_2) = \frac12\left(\frac57 + \frac25\right) = \frac{39}{70},$$ and $$P(W_2 \mid W_1) = P(W_2 \mid W_1, U_1)P(U_1) + P(W_2 \mid W_1, U_2)P(U_2) = \frac12\left(\frac12 + \frac34\right) = \frac58.$$ Then $p = 39/112$.

I am confident that the first factor is correct, so I suppose the error lies in the second one?

$\endgroup$
  • $\begingroup$ Where is the formula $$P(W_2 \mid W_1) = P(W_2 \mid W_1, U_1)P(U_1) + P(W_2 \mid W_1, U_2)P(U_2) $$ supposed to come from? $\endgroup$ – Did Jul 17 '16 at 15:10
  • $\begingroup$ @Did From my intuition. I didn't know how to write a probability conditioned by two events, hence the title. $\endgroup$ – rubik Jul 17 '16 at 15:11
  • $\begingroup$ My suggestion would be to review Bayes formula in depth. $\endgroup$ – Did Jul 17 '16 at 15:12
2
$\begingroup$

In your solution, $W_1$ is actually ambiguous---$P(W_2|W_1)$ varies depending on the urn you first chose. Thus you can't write $P(W_1\text{ and }W_2) = P(W_1)P(W_2|W_1)$.

Instead, you should separate into the two cases and add at the end, to obtain a probability of $$ \frac 12\left(\frac 57\cdot\frac 12+\frac 25\cdot\frac 34\right) = \frac{23}{70}. $$

$\endgroup$
  • $\begingroup$ I don't understand how the Bayes formula would be applied in this case. I get $$P(W_2 \mid W_1) = \frac{P(W_1 \mid W_2)P(W_2)}{P(W_1)}$$ but $P(W_1 \mid W_2)$ doesn't make sense! $\endgroup$ – rubik Jul 17 '16 at 15:17
  • $\begingroup$ A comment on Bayes' rule: $P(W_1|W_2)$ makes sense in that it computes the probability that the first ball was white, if you walked in late and only got to see the drawing of the second ball. $\endgroup$ – cubesteak Jul 17 '16 at 15:22
  • $\begingroup$ And a comment on Bayes' rule for this problem: You want to write $P(W_2\text{ and }W_1|U_1) = P(W_2|W_1,U_1)P(W_1|U_1)$. $\endgroup$ – cubesteak Jul 17 '16 at 15:28
2
$\begingroup$

Split it into disjoint events, and then add up their probabilities:

  • $P(w_1,w_2)=\frac12\cdot\frac{5}{2+5}\cdot\frac{2+1}{3+2+1}=\frac{5}{28}$
  • $P(w_2,w_1)=\frac12\cdot\frac{2}{3+2}\cdot\frac{5+1}{2+5+1}=\frac{3}{20}$

The overall probability is therefore $\frac{5}{28}+\frac{3}{20}=\frac{23}{70}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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