# Probability conditioned by two events

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?

• 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? – Did Jul 17 '16 at 15:10
• @Did From my intuition. I didn't know how to write a probability conditioned by two events, hence the title. – rubik Jul 17 '16 at 15:11
• My suggestion would be to review Bayes formula in depth. – Did Jul 17 '16 at 15:12

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 $$\frac 12\left(\frac 57\cdot\frac 12+\frac 25\cdot\frac 34\right) = \frac{23}{70}.$$

• 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! – rubik Jul 17 '16 at 15:17
• 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. – cubesteak Jul 17 '16 at 15:22
• And a comment on Bayes' rule for this problem: You want to write $P(W_2\text{ and }W_1|U_1) = P(W_2|W_1,U_1)P(W_1|U_1)$. – cubesteak Jul 17 '16 at 15:28

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 $\frac{5}{28}+\frac{3}{20}=\frac{23}{70}$