Understanding the Urn Probability Problem

This question asks for the probability that an urn containing 5 balls is composed entirely of white balls, given that we have drawn two balls from it and both were found to be white. This is a classic problem involving conditional probability and can be solved using Bayes' Theorem or by considering the probabilities of observing the result under different possible initial states of the urn.

The observation that the first two balls drawn are white is important evidence that changes our belief about the initial composition of the urn. We need to update our probability based on this new information.

Defining Events and Hypotheses

Let's define the relevant events and the possible initial states (hypotheses) for the urn:

• Event B: The first two balls drawn from the urn are white.
• Hypothesis \(C_k\): The urn initially contains \(k\) white balls and \(5-k\) black balls, for \(k \in \{0, 1, 2, 3, 4, 5\}\).

We are interested in finding the probability that the urn contains only white balls, given that the first two draws were white. The event that the urn contains only white balls corresponds to the hypothesis \(C_5\) (5 white, 0 black). So, we want to calculate \(P(C_5|B)\).

Since we observed two white balls drawn, the initial composition of the urn must have contained at least 2 white balls. Therefore, the possible initial hypotheses compatible with the observation B are \(C_2, C_3, C_4, C_5\).

Prior Probabilities of Compositions

The problem does not provide information about the prior probabilities of the different urn compositions. A standard approach in such cases, especially when the options suggest a particular interpretation, is to assume that the possible initial compositions that could produce the observed result are equally likely. The compositions that can produce two white balls are \(C_2, C_3, C_4, C_5\).

So, we assume the prior probabilities for these relevant compositions are equal:

\[P(C_2) = P(C_3) = P(C_4) = P(C_5) = \frac{1}{4}\]

We assume \(P(C_0) = P(C_1) = 0\) in this context because these compositions cannot result in drawing two white balls.

Calculating Likelihoods: Probability of Drawing Two White Balls Given Composition

Now, let's calculate the likelihood \(P(B|C_k)\), which is the probability of drawing two white balls given that the urn has composition \(C_k\). We assume the balls are drawn without replacement. The total number of ways to draw 2 balls from 5 is \(\binom{5}{2}\). The number of ways to draw 2 white balls from \(k\) white balls is \(\binom{k}{2}\).

The probability \(P(B|C_k)\) for \(k \ge 2\) is given by:

\[P(B|C_k) = \frac{\text{Number of ways to draw 2 white balls}}{\text{Total number of ways to draw 2 balls}} = \frac{\binom{k}{2}}{\binom{5}{2}}\]

We calculate this for each relevant composition (\(k=2, 3, 4, 5\)):

• For \(C_2\) (2 White, 3 Black): \(P(B|C_2) = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}\).
• For \(C_3\) (3 White, 2 Black): \(P(B|C_3) = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10}\).
• For \(C_4\) (4 White, 1 Black): \(P(B|C_4) = \frac{\binom{4}{2}}{\binom{5}{2}} = \frac{6}{10}\).
• For \(C_5\) (5 White, 0 Black): \(P(B|C_5) = \frac{\binom{5}{2}}{\binom{5}{2}} = \frac{10}{10} = 1\). If the urn is all white, drawing two white balls is certain.

Here is a summary:

Calculating Total Probability of Observing Two White Balls

To use Bayes' Theorem, we need the total probability of the observed event B, which is \(P(B)\). We can calculate this using the law of total probability, summing over all possible compositions that could lead to B:

\[P(B) = \sum_{k=2}^{5} P(B|C_k) P(C_k)\]

Using the prior probabilities and likelihoods calculated above:

\[P(B) = P(B|C_2)P(C_2) + P(B|C_3)P(C_3) + P(B|C_4)P(C_4) + P(B|C_5)P(C_5)\]

\[P(B) = \left(\frac{1}{10} \times \frac{1}{4}\right) + \left(\frac{3}{10} \times \frac{1}{4}\right) + \left(\frac{6}{10} \times \frac{1}{4}\right) + \left(1 \times \frac{1}{4}\right)\]

\[P(B) = \frac{1}{40} + \frac{3}{40} + \frac{6}{40} + \frac{10}{40} = \frac{1+3+6+10}{40} = \frac{20}{40} = \frac{1}{2}\]

Applying Bayes' Theorem to Find the Posterior Probability

We want to find \(P(C_5|B)\), the probability that the urn is all white (\(C_5\)) given that the first two draws were white (B). Bayes' Theorem states:

\[P(C_5|B) = \frac{P(B|C_5) P(C_5)}{P(B)}\]

Substitute the values we calculated:

• Likelihood of B given \(C_5\): \(P(B|C_5) = 1\)
• Prior probability of \(C_5\): \(P(C_5) = 1/4\)
• Total probability of B: \(P(B) = 1/2\)

\[P(C_5|B) = \frac{1 \times \frac{1}{4}}{\frac{1}{2}}\]

\[P(C_5|B) = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{1} = \frac{2}{4} = \frac{1}{2}\]

Final Answer

Based on the calculation using Bayes' Theorem and assuming equal prior probabilities for the possible compositions that could yield two white balls, the probability that the urn contains only white balls given the first two draws were white is \(1/2\).

Revision Table: Probability and Bayesian Inference

Additional Information: Interpreting Prior Probabilities in Urn Problems

The choice of prior probabilities \(P(C_k)\) is crucial in Bayesian problems. In this solution, we assumed that among the compositions from which two white balls could be drawn (\(C_2, C_3, C_4, C_5\)), each was equally likely initially. This is one common way to handle unspecified priors in textbook problems, often leading to one of the provided answers.

Another common assumption, especially in introductory contexts, is that *all* possible compositions (from 0 white, 5 black to 5 white, 0 black) are equally likely a priori (\(P(C_k) = 1/6\) for \(k=0, \dots, 5\)). If we used this prior, the calculation for \(P(B)\) would be \(\sum_{k=0}^5 P(B|C_k) P(C_k) = \frac{1}{6} \sum_{k=0}^5 P(B|C_k)\). Since \(P(B|C_0)=0\) and \(P(B|C_1)=0\), this sum would be \(\frac{1}{6} (P(B|C_2) + P(B|C_3) + P(B|C_4) + P(B|C_5)) = \frac{1}{6}(\frac{1}{10} + \frac{3}{10} + \frac{6}{10} + 1) = \frac{1}{6}(\frac{20}{10}) = \frac{2}{6} = \frac{1}{3}\).

Then \(P(C_5|B) = \frac{P(B|C_5) P(C_5)}{P(B)} = \frac{1 \times \frac{1}{6}}{\frac{1}{3}} = \frac{1/6}{1/3} = \frac{1}{2}\).

Interestingly, both the prior assumption of equal likelihood among compatible compositions (\(C_2\) to \(C_5\)) and the prior assumption of equal likelihood among all possible compositions (\(C_0\) to \(C_5\)) lead to the same posterior probability of \(1/2\) for \(C_5\) given B. This consistency reinforces confidence in the result for this specific problem structure.