1. **Problem statement:** An urn contains 5 balls. Two balls are drawn and both are white. We want to find the probability that all 5 balls in the urn are white. 2. **Understanding the problem:** We have partial information from the draw (2 white balls). We want to find the probability that all 5 balls are white given this information. 3. **Key concept:** Use conditional probability and Bayes' theorem. Let $A$ be the event "all 5 balls are white" and $B$ be the event "2 drawn balls are white". 4. We want $P(A|B)$, the probability all balls are white given the 2 drawn are white. 5. By Bayes' theorem: $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)} $$ 6. Assume prior probability $P(A)$ is unknown or uniform. For simplicity, assume $P(A) = p$ and $P(\text{not } A) = 1-p$. 7. Calculate $P(B|A)$: If all 5 balls are white, drawing 2 balls will always be white, so $$ P(B|A) = 1 $$ 8. Calculate $P(B|\text{not } A)$: If not all balls are white, at least one ball is not white. Without more info, assume the probability of drawing 2 white balls is less than 1, say $q$. 9. Total probability of drawing 2 white balls: $$ P(B) = P(B|A)P(A) + P(B|\text{not } A)P(\text{not } A) = 1 \cdot p + q \cdot (1-p) = p + q(1-p) $$ 10. Therefore, $$ P(A|B) = \frac{p}{p + q(1-p)} $$ 11. Without prior $p$ and $q$, we cannot find a numeric answer. If we assume prior $p$ is very small and $q$ is small, $P(A|B)$ increases but remains less than 1. **Summary:** The probability all balls are white given 2 drawn are white depends on prior assumptions. If you assume all balls are equally likely to be white or not, and no other info, the problem is incomplete for a numeric answer. **If you assume the urn contains only white and non-white balls and the prior probability of all white is uniform, then the probability that all balls are white given 2 drawn white is** $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{1 \cdot \frac{1}{\binom{5}{5}}}{P(B)} = \text{depends on } P(B) $$ **Hence, more information is needed for a numeric answer.**