(Urn problem) What is the probability the first ball drawn was red, given that the second ball was black?

Question

The following question is taken from Mark Joshi's quant job interview.

Urn A contains 5 white balls and 2 red balls. Urn B contains 4 red balls and 4 black balls. You randomly pick an urn and draw a ball (without replacement). You then repeat the process of selecting an urn and drawing out a ball. What is the probability the first ball drawn was red, given that the second ball was black?

This looks like a Bayesian question. Let $X_1$ and $X_2$ be the colour of ball picked in first and second respectively. Then $$P(X_1 = red | X_2 = Black) = \frac{P(X_2 = Black | X_1 = red) P(X_1 = red)}{P(X_2 = Black)}.$$ Note that $$P(X_1 = red) = \frac{1}{2}\times \frac{4}{8} \quad \text{and} \quad P(X_2 = Black | X_1 = red) = \frac{1}{2}\times \frac{2}{7} + \frac{1}{2} \times \frac{4}{7}.$$ However, I do not know how to calculate $$P(X_2 = Black).$$

Answer 1

By definition, the desired conditional probability is:

$$\Pr(X_1 = \text{red} \mid X_2 = \text{Black}) = \frac{ \Pr\{\text{$X_1$ is red and $X_2$ is black}\} }{ \Pr\{X_2 ~\text{is black}\} } ~.$$

$\renewcommand{\bg}[1]{ \bigl( #1 \bigr) }$I shall denote the outcomes of $(X_1,X_2)$ using "$\text{URN} \otimes {\text{color}}$". For example, $(B \otimes r,A \otimes w)$ means the drawing a red ball from urn $B$ first, then getting a white ball from urn $A$.

It's not explicitly specified in the question statement, but presumably randomly pick an urn means the two drawings are independent and with identical probability $p = 1/2$ of selecting urn $A$. Namely, the chance of picking urn $B$ is $1-p$.

The event in the denominator yields: $$\{X_2 ~\text{is black}\} = \left\{\begin{aligned} &~~(A \otimes w, B \otimes b) \\ \text{or} & ~~(A \otimes r, B \otimes b) \\ \text{or} & ~~(B \otimes r, B \otimes b) \\ \text{or} &~~(B \otimes b, B \otimes b)\end{aligned} \right. \implies \Pr\{X_2 ~\text{is black}\} = \left\{\begin{aligned} &\hphantom{{}+{}}p \frac57\cdot (1-p)\frac48 \\ & + p \frac27\cdot (1-p)\frac48 \\ & + (1-p) \frac48\cdot (1-p)\frac47 \\ & + (1-p) \frac48\cdot (1-p)\frac37\end{aligned} \right.\\ \implies \Pr\{X_2 ~\text{is black}\} = \frac{1-p}{56}\bg{ 20p+ \color{red}{8p+16(1-p)}+12(1-p)} = \frac{1-p}2 = \frac14$$

The event in the numerator is (in fact the $\color{red}{\text{middle two terms}}$ of the denominator): $$\{\text{$X_1$ is red and $X_2$ is black}\} = \left\{\begin{aligned} & ~~(A \otimes r, B \otimes b) \\ \text{or} & ~~(B \otimes r, B \otimes b) \end{aligned} \right. \\ \implies \Pr\{\text{$X_1$ is red and $X_2$ is black}\} = \frac{1-p}{56}\bg{ \color{red}{8p+16(1-p)}} = \frac{(1-p)(2-p)}7 = \frac3{28}$$ Notice that at $p = \frac{1-p}2 = \frac12$ one only needs to look at the coefficients to calculate the desired conditional probability: $$\frac{ \Pr\{\text{$X_1$ is red and $X_2$ is black}\} }{ \Pr\{X_2 ~\text{is black}\} } = \frac{ 8+16 }{ 20+8+16+12 } = \frac{24}{56} = \frac37$$ Formally, it is $$\frac{ \Pr\{\text{$X_1$ is red and $X_2$ is black}\} }{ \Pr\{X_2 ~\text{is black}\} } = \frac{(1-p)(2-p)/7}{(1-p)/2} = \frac{2(2-p)}7 = \frac{ 3/28 }{ 1/4 } = \frac37$$