A bag initially contains three unknown balls. We add one white ball, then draw one ball at random. If all hypotheses about the initial composition (0,1,2,3 whites) are equally likely, what is the probability that the drawn ball is white?

Introduction / Context:
This problem uses a simple Bayesian averaging over hypotheses. Before adding a white ball, the bag has three balls but we do not know how many are white. We assume four hypotheses are equally likely: initially 0, 1, 2, or 3 whites. After adding one white ball, a single draw is made; we want the overall chance the drawn ball is white.

Given Data / Assumptions:

• Prior over k ∈ {0,1,2,3} whites initially: each 1/4.
• After adding one white, the bag has 4 balls with (k+1) white.
• On a single draw: P(white | k) = (k+1)/4.

Concept / Approach:
Law of total probability: average the conditional probabilities over the four equally likely hypotheses.

Step-by-Step Solution:

Verification / Alternative check:
Explicit mixture: If initially 0W ⇒ 1/4; 1W ⇒ 2/4; 2W ⇒ 3/4; 3W ⇒ 4/4; average of {1/4, 1/2, 3/4, 1} is (2.5)/4 = 0.625 = 5/8.

Why Other Options Are Wrong:
3/4 (0.75) would require biasing toward larger k; 1/2 (0.5) or 3/8 (0.375) undervalue the influence of adding the extra white ball.

Common Pitfalls:
Assuming the initial composition is uniformly random over individual colorings rather than over k; the statement says “all hypotheses concerning the colour” by count, i.e., each k equally likely.

Final Answer:
5/8