0
$\begingroup$

I have an intuitive answer here, but I struggle setting up probability questions, so any pointers would be great.

There are 2 urns of 50 balls: one with 50 red balls and another with 49 red and one green. What is the probability of drawing a red conditional on drawing 49 reds from a randomly-sampled urn.

Here's how I set it up:

Call the urns $U_{1}$ and ${U_{2}}$, where $P(U_{1}) = P(U_{2}) = \frac{1}{2}$. Label drawing $n = 49$ consecutive red balls first event $A$, not drawing 49 red balls $\neg A$, and drawing a ball event $B = \{r, g\}$.

We are interested in $P(B = r \mid A)$, the probability of drawing a red ball after the event of drawing 49 consecutive reds from a randomly sampled urn. Using Baye's rule:

$$ P(B = r \mid A) = \frac{P(A \mid B = r) P(B = r)}{P(A)} $$

The probability of drawing 49 reds given we've drawn a red ball is equal to the probability of drawing the all-red urn.

$$ P(A \mid B = r) = P(A \mid B = r, U_{1}) P(U_{1}) + P(A \mid B = r) P(U_{2}) = 1 \frac{1}{2} + 0 \frac{1}{2} = \frac{1}{2} $$

The probability of a red ball overall requires summing over the possible ways of getting a red ball, i.e. a red ball from urn 1 or urn 2 and a red ball after sampling 49 red balls or not:

$$ \begin{align} P(R) &= P(B = r, A, U_{1}) + P(B = r, \neg A, U_{1}) + P(B = r, A, U_{2}) + P(B = r, \neg A, U_{2})\\ &= \frac{1}{2} + 0 + 0 + \frac{49}{50} (1 - \frac{49!}{50!}) \frac{1}{2}\\ &= 0.9802\\ \end{align} $$

Finally, the probability of 49 reds also requires us to marginalize over the possible ways of getting 49 reds:

$$ \begin{align} P(A) & = P(A, B = r, U_1) + P(A, B = g, U_1) + P(A, B = r, U_2) + P(A, B = g, U_2)\\ &= \frac{1}{2} + 0 + 0 + \frac{1}{50} \times \frac{1}{2}\\ &= 0.51 \end{align} $$

Plugging these values into Bayes' rule gives me $0.961$.

Can anyone help with whether that is correct, and whether there are any better ways to set it up?

$\endgroup$
1
  • $\begingroup$ To be clear: you select an urn uniformly at random and extract $49$ balls from it without replacement, note that all of those were red and then ask for the probability that the last ball in the urn is also red? $\endgroup$
    – lulu
    Commented Sep 4 at 14:32

3 Answers 3

Reset to default
1
$\begingroup$

Note: You do not need $B$ to be a random variable, when you may make it an event, just like $A$.

So let $A$ be the event of drawing 49 red balls, $B$ be the event of the remaining ball being red, and $U_1, U_2$ the events for drawing from the indicated urn.

Of interest: when drawing from Urn 1, $A$ and $B$ are both certain (all balls must be red), while when drawing from Urn 2, $A$ and $B$ are complements, with $\mathsf P(A\mid U_2)=1/50$. So:

$$\begin{align}\mathsf P(B\mid A) &=\dfrac{\mathsf P(A, B)}{\mathsf P(A)}\\&= \dfrac{\mathsf P(A, B\mid U_1)\mathsf P(U_1)+\mathsf P(A, B\mid U_2)\mathsf P(U_2)}{\mathsf P(A\mid U_1)\mathsf P(U_1)+\mathsf P(A\mid U_2)\mathsf P(U_2)}\\&=\dfrac{1+0}{1+1/50}\\&= \dfrac{50}{51}\end{align}$$

$\endgroup$
7
  • $\begingroup$ Thank you. When I tried to compute $P(A \mid B)$ and $P(B)$, separately, I ended up $P(A \mid B) = 1/2$, but that wasn't correct. $P(A \mid B) = \sum_{k=1}^{2} P(A, U_{k} \mid B) P(U_k \mid B) = 1 + 0$. This means $P(B) = 0.5$. My confusion still is that, if $P(B) = \sum_{k=1}^{2} P(B, A, U_k) = 0.5$, then how do we write out $P(A)$ -- to me we need to write $P(A) = \sum_{k=1}^{2} \sum_{b=r, g} P(A, B_b, U_k)$ to get $1 + 1/50$. But why vary the outcomes of $B$ in this summation, but don't vary the possible outcomes of the first 49 balls ($A$ v $\neg A$) in $P(B)$? This is what led me astray. $\endgroup$
    – user_15
    Commented Sep 5 at 11:18
  • $\begingroup$ @user_15 Well, firstly, $\mathsf P(A\mid B)=\sum_{k=1}^2\mathsf P(A\mid U_k,B)\,\mathsf P(U_k\mid B)$ which equals $50/99$ . Secondly, you only need to partition over the urns. $\mathsf P(B) = \sum_{k=1}^2 \mathsf P(B, U_k)$ equalling $99/100$ and likewise $\mathsf P(A) = \sum_{k=1}^2\mathsf P(A, U_k)$ equalling $51/100$ $\endgroup$
    – Graham Kemp
    Commented Sep 5 at 12:17
  • $\begingroup$ Thanks, again. Is it correct to say, then, that an event is a discrete random variable when there are multiple options? In this example, $A$ and $B$ are events and do not need to made a d.r.v, but $U$ is a random variable because it can take on different outcomes -- $U_1$ and $U_2$? This whole time I've been saying there's three events and, when it comes to computing marginal probabilities, like $P(B)$, I keep trying to shoe-horn the other two events in there, which is leading to problems. $\endgroup$
    – user_15
    Commented Sep 5 at 12:47
  • 1
    $\begingroup$ @user_15 The point is only to do that which is useful. Partitioning over the urns is useful because having, or not, a green ball in the mix allows you to easily evaluate the conditional probabilities.$$\def\P{\mathsf P}\boxed{\begin{array}{c|c}\P(A\mid U_1)=1&\P(B\mid U_1)=1\\\hline\P(A\mid U_2)=1/50&\P(B\mid U_2)=49/50\end{array}}$$Now you *could* go further, but how useful is it?$$\boxed{\begin{array}{c|c}\P(A\mid U_1, B)=1&\P(B\mid U_1,A)=1&\P(A\mid U_1,\neg B)=0&\P(B\mid U_1,\neg A)=0\\\hline\P(A\mid U_2, B)=0&\P(B\mid U_2, A)=0&\P(A\mid U_2,\neg B)=1&\P(B\mid U_2,\neg A)=1\end{array}}$$ $\endgroup$
    – Graham Kemp
    Commented Sep 5 at 23:12
  • 1
    $\begingroup$ $\begin{align}\mathsf P(A)&=\sum_{k=1}^2\sum_{\beta\in\{B,\neg B\}}\mathsf P(U_k)\mathsf P(\beta\mid U_k)\mathsf P(A\mid U_k,\beta)\\&=\tfrac12\cdot 1\cdot 1+\tfrac12\cdot0\cdot0+\tfrac 12\cdot\tfrac{49}{50}\cdot0+\tfrac 12\cdot\tfrac 1{50}\cdot1\\&=\tfrac{51}{100}\end{align}$ $\endgroup$
    – Graham Kemp
    Commented Sep 5 at 23:27
1
$\begingroup$

$$P(B=r|A)=P(U1|A)=\frac{P(U1 \cap A)}{P(A)}$$ Note that the numerator is $P(U1)$ and the denominator $$P(U1 \cap A)+P(U2 \cap A)=P(U1)+P(U2)P(A|U2)$$ now by symmetry this last term is equal to $1/50$ or the chance that the green ball appears in the last position of a random drawing order, since you have given that $P(U1)=P(U2)$ the terms cancel and we are left with $$P(B=r|A)=\frac{1}{1+P(A|U2)}=\frac{50}{51}$$ Which should be slightly more than 0.98.

$\endgroup$
6
  • $\begingroup$ Thanks! Intuitively, this makes sense. But is there a way to arrive at this answer by not equating $P(B=r \mid A) = P(U_{1} \mid A)$? i.e. by keeping all events as I did in my OP. I'm trying to be better at setting up these problems, and it would be nice to see why I am not getting the same answer. $\endgroup$
    – user_15
    Commented Sep 4 at 15:53
  • 1
    $\begingroup$ @user_15 Your error is the evaluation of $\mathsf P(B=r,\neg A, U_2)$ in the calculation for $\mathsf P(R)$. $$\begin{align}\mathsf P(B=r,\neg A,U_2) &=\mathsf P(B=r,\neg A\mid U_2)\,\mathsf P(U_2)\\& = \tfrac {49}{50}\,\tfrac 12\\&=\tfrac {49}{100}\end{align}$$ Since when drawing from the second urn, $B=r$ and $\neg A$ are the same event (that the green ball is among the first 49). $\endgroup$
    – Graham Kemp
    Commented Sep 5 at 3:14
  • $\begingroup$ @user_15 by using sigma algebra's countable additivity $P(B|A)=(P(B,U1,A)+P(B,U2,A))/P(A)$ the second therm is the intersection of all events where 49 red balls are present, the first 49 drawn are red and the last drawn is red. This provides a contradiction since no order of 49 red balls and 1 green ball contains 50 red balls. As for the first term $U1\subset B$ since every order of 50 red balls contains a red ball at the fiftieth position. An analogous argument is used for $P(U1\cap A)=P(U1)$, in fact just $A\cap B=U1$ suffices to solve this, but is not as general as the stepwise approach. $\endgroup$
    – jucom
    Commented Sep 5 at 16:46
  • $\begingroup$ As a general tip if you don't want to use intuitive arguments always try to get a picture of what the events involved are, and then work towards events you have evaluations for using intersections and partitions. $\endgroup$
    – jucom
    Commented Sep 5 at 16:56
  • $\begingroup$ Thanks @jucom. In the thread above, Graham Kemp sorted my marginalization issue out. Intuitively, I understand why the events are collapsed into $A \cap B = U1$, I was just making a mistake when marginalizing across the three events. $\endgroup$
    – user_15
    Commented Sep 6 at 8:34
1
$\begingroup$

The way your question is framed, it is not clear that conditionality is involved.

On the other hand, using your symbols $U_1,U_2$ and $49R$ for $49$ consecutive reds if you are asking $P(U_1\mid 49R)$, it becomes

$\dfrac {P(U_1\cap49R)}{P(U_1\cap49R) + P(U_2\cap49R)}$

= $\dfrac{\frac12\cdot1}{\frac12\cdot1 + \frac12\cdot\frac{1}{50}} = \frac{50}{51}$

$P(U_2 \mid 49R)$ will obviously be the complement, $1 - \frac{50}{51} =\frac{1}{51}$

Three hints that you may find useful are

  • see if you can reduce it to a familiar class of problems, like supplies from two factories with differing OK rates
  • use as few mathematical symbols as essential (can even use some English for greater clarity, but only where it can't be misinterpreted)
  • Use the "baby" form of Bayes' Rule until you gather more expertise
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .