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 $$P(X_2 = Black).$$

  • $\begingroup$ You overlooked the possibility that a red ball could be selected from Urn A in the first selection. $\endgroup$ Commented Feb 15, 2020 at 11:56
  • $\begingroup$ @N.F.Taussig More precisely which probability are you discussing? $\endgroup$
    – Idonknow
    Commented Feb 15, 2020 at 12:01
  • 2
    $\begingroup$ As well as the colour of the balls picked, you need to consider which urn they were selected from in each round. $\endgroup$ Commented Feb 15, 2020 at 12:08
  • $\begingroup$ Let $Y_1,Y_2$ identify which urn was selected each round. $$\mathsf P(X_1=\textsf{red}, Y_1=\textsf{urnA})=\tfrac 12\times \tfrac 27\\\mathsf P(X_1=\textsf{red}, Y_1=\textsf{urnB})=\tfrac 12\times \tfrac 48$$, et cetera $\endgroup$ Commented Feb 15, 2020 at 12:21

1 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 $$\frac{ \Pr\{\text{$X_1$ is red and $X_2$ is black}\} }{ \Pr\{X_2 ~\text{is black}\} } = \frac{(1-p)(2-p)/7}{(1-p)/2} = \frac{2(2-p)}7 = \frac{ 3/28 }{ 1/4 } = \frac37$$


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