Question

An urn contains \(b\) black balls and \(r\) red balls. One of the balls is drawn at random, but when it is put back in the urn \(c\) additional balls of the same color are put in with it. Now suppose that we draw another ball. Show that the probability that the first ball is drawn was black given that the second ball drawn was red is \(b /(b+r+c)\).

Short answer

The probability that the first ball drawn was black given that the second ball drawn was red is \(\frac{b}{b+r+c}\).

Step-by-step solution

Probabilities of drawing a red ball second

If the first ball was black (B1), the probability of drawing a red ball (R2) is: \(P(R_{2} \mid B_{1}) = \frac{r}{(b+c)+r}\) If the first ball was red (R1), the probability of drawing a red ball (R2) is: \(P(R_{2} \mid R_{1}) = \frac{(r-1+c)}{(b-1)+r+c}\) \( \) Now, we can use Bayes' theorem to find the probability that the first ball drawn was black, given that the second ball drawn is red: \( \)

Bayes' theorem

\(P(B_{1} \mid R_{2}) = \frac{P(R_{2} \mid B_{1}) \cdot P(B_{1})}{P(R_{2} \mid B_{1}) \cdot P(B_{1}) + P(R_{2} \mid R_{1}) \cdot P(R_{1})}\) We know \(P(B_{1}) = \frac{b}{b+r}\) and \(P(R_{1}) = \frac{r}{b+r}\). Substituting the values: \(P(B_{1} \mid R_{2}) = \frac{\frac{r}{(b+c)+r} \cdot \frac{b}{b+r}}{\frac{r}{(b+c)+r} \cdot \frac{b}{b+r} + \frac{(r-1+c)}{(b-1)+r+c} \cdot \frac{r}{b+r}}\) \( \) Simplify the expression to find the probability that the first ball drawn was black, given that the second ball drawn is red: \( \)

Simplify the expression

\(P(B_{1} \mid R_{2}) = \frac{rb}{rb + (r-1+c)r} = \frac{b}{b+r+c}\) \( \) Now we have shown that the probability that the first ball is drawn was black given that the second ball drawn is red is \(b /(b+r+c)\).

Key concepts

Bayes' theorem

Bayes' theorem is a fundamental concept in the field of probability. It allows us to update our beliefs about the likelihood of an event based on new evidence. In the context of the urn problem discussed, we're interested in reversing the conditional probability — that is, we know the color of the second ball and want to find out the probability regarding the color of the first ball drawn.

Mathematically, Bayes' theorem is expressed as:
\[\begin{equation} P(A | B) = \frac{P(B | A) \times P(A)}{P(B)}\end{equation}\]
Where:

• P(A | B) is the probability of event A occurring given that B is true.
• P(B | A) is the probability of event B given event A is true.
• P(A) is the probability of event A.
• P(B) is the total probability of event B happening.

The theorem allows us to adjust prior probability estimates, P(A), in light of the probabilities of observed events, P(B | A) and P(B). In the exercise, after using Bayes' theorem we find that the probability the first ball was black given the second ball drawn is red is \[\begin{equation} \frac{b}{b + r + c}\end{equation}\], which is a classic application of Bayes' theorem in solving conditional probability problems.

Probability Models

Probability models are mathematical representations of random processes. They are used to describe complex scenarios and find probabilities of different outcomes. In our brain-teasing urn problem, we actually have a simple probability model at play.

• Sample Space: The total number of possible outcomes in our experiment - drawing a ball from the urn.
• Events: Specific outcomes of interest, such as drawing a black or red ball first or second.
• Probabilities: Numbers assigned to these events that represent how likely they are to occur.

The urn problem demonstrates how probability models can change after an event occurs - in this case, the inclusion of additional colored balls. We adjust our model after each draw, considering the incremented number of balls. This dynamic change is common in real-life scenarios, where prior events affect the likelihood of future events. Breaking down the process into a structured probability model helps in systematically approaching the problem and applying tools like Bayes' theorem effectively.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations, arrangements, and counting. It plays a key role in computing probabilities when we need to determine the number of ways certain events can occur.

Although not directly mentioned in the provided solution, combinatorics is integral to understanding the underlying principles of probability models. For example, in our urn scenario, the number of ways of drawing balls would form the basis for determining the sample space. Here's why combinatorics is important:

• It helps to count without actually enumerating all possible outcomes.
• It forms the foundation for calculating probabilities in scenarios with multiple outcomes.

In many probability problems, especially those involving a series of events (like multiple draws from an urn), combinatorial aspects such as permutations and combinations are used to calculate probabilities. Although our urn problem is relatively straightforward, more complex scenarios might require intricate combinatorial reasoning to account for all possible outcomes and their associated probabilities.