The question asks about the probability of exactly one person out of three having a limiting long-term illness or disability in Wales. In 2001, a quarter of the population of Wales reported having such an illness or disability that restricted their daily activities.

To solve this problem, we can use the binomial probability formula. The binomial probability formula is used to find the probability of getting exactly βkβ successes out of βnβ trials, given a success probability of βpβ. In this case, the success is a person reporting a limiting long-term illness or disability, and the failure is a person not reporting such an illness or disability.

Letβs assume that the probability of a person reporting a limiting long-term illness or disability is βpβ = 1/4, and the probability of a person not reporting such an illness or disability is βqβ = 1 - p = 3/4. The number of trials is βnβ = 3, and the number of successes we want is βkβ = 1.

Using the binomial probability formula, the probability of exactly one person out of three having a limiting long-term illness or disability is given by:

P(X = 1) = {n \choose k} \cdot p^k \cdot q^{n-k}

where {n \choose k} is the binomial coefficient, also known as βn choose kβ. The binomial coefficient represents the number of ways to choose βkβ items from a set of βnβ items, and is given by:

{n \choose k} = \frac{n!}{k! (n-k)!}

Plugging in the values for βnβ, βkβ, βpβ, and βqβ, we get:

P(X = 1) = {3 \choose 1} \cdot \left(\frac{1}{4}\right)^1 \cdot \left(\frac{3}{4}\right)^2 = 3 \cdot \frac{1}{4} \cdot \frac{9}{16} = \frac{27}{64}

So, the probability of exactly one person out of three having a limiting long-term illness or disability in Wales is \frac{27}{64}, which is the answer choice B.