This problem is an example of Bayesian probability, where we want to update our belief about the probability that all the balls in the urn are white, given the evidence that two balls drawn are white.

Let’s define the problem:

• Let $H_1$: All 5 balls in the urn are white.
• Let $H_2$: Not all the balls in the urn are white (i.e., some other colors are present).
• The evidence $E$: Two balls are drawn, and both are white.

We are interested in the posterior probability $P(H_1 \mid E)$, i.e., the probability that all the balls are white given that two white balls were drawn.

We can use Bayes’ theorem for this: $P(H_1 \mid E) = \frac{P(E \mid H_1) P(H_1)}{P(E)}$

Step 1: Calculate $P(E \mid H_1)$

If all the balls in the urn are white (i.e., $H_1$), then the probability of drawing two white balls is 1, because every ball is white: $P(E \mid H_1) = 1$

Step 2: Calculate $P(E \mid H_2)$

If not all the balls in the urn are white (i.e., $H_2$), we need more information about the distribution of the other balls. For now, let's assume that the urn has at least one ball that is not white.

The probability $P(E \mid H_2)$ will depend on the specific composition of the non-white balls, but for simplicity, we assume that there's an equal chance that the drawn balls could be non-white, so: $P(E \mid H_2) = \text{some probability less than 1}$

Step 3: Assign prior probabilities $P(H_1)$ and $P(H_2)$

Without any other information, we might assign equal prior probabilities: $P(H_1) = P(H_2) = \frac{1}{2}$

Step 4: Calculate $P(E)$

$P(E)$ is the total probability of drawing two white balls, which is a weighted average: $P(E) = P(E \mid H_1) P(H_1) + P(E \mid H_2) P(H_2)$ Since $P(E \mid H_1) = 1$, we get: $P(E) = P(H_1) + P(E \mid H_2) P(H_2)$

Step 5: Apply Bayes' theorem

Now, applying Bayes' theorem: $P(H_1 \mid E) = \frac{P(E \mid H_1) P(H_1)}{P(H_1) + P(E \mid H_2) P(H_2)}$ Substituting values: $P(H_1 \mid E) = \frac{1 \times \frac{1}{2}}{\frac{1}{2} + P(E \mid H_2) \times \frac{1}{2}}$ Simplifying: $P(H_1 \mid E) = \frac{1}{1 + P(E \mid H_2)}$

Final Answer:

The probability that all the balls are white depends on $P(E \mid H_2)$, which represents the likelihood of drawing two white balls if not all the balls are white. If we had more information about the urn's composition under $H_2$, we could calculate this precisely.

Do you want more details on how to approach this problem, or any follow-up questions?

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Here are 5 related questions for further exploration:

1. How does the assumption of equal prior probabilities affect the result?
2. What if we knew the exact number of non-white balls in the urn?
3. How would the result change if 3 balls were drawn instead of 2?
4. Can this problem be solved without Bayesian methods?
5. How does the problem change if we replace white balls with differently colored balls?

Tip: In Bayesian problems, having prior information (or making justified assumptions) greatly impacts the outcome. Always consider how priors influence the posterior probability.