# Probability BMAT 2004 Q16

In 2001 a quarter of the population of Wales reported having a limiting long-term illness or
disability which restricted their daily activities.
What is the probability that out of a random group of three people who live in Wales,
exactly one of them will have reported such an illness or disability?

A 9/64
B 27/64
C 37/64
D 3/4

Hi everyone!
The answer is B), iβm not sure how to get to that solution
Thanks!

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 βkβ successes out of βnβ trials, given a success probability of βpβ. 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.

Letβs assume that the probability of a person reporting a limiting long-term illness or disability is βpβ = 1/4, and the probability of a person not reporting such an illness or disability is βqβ = 1 - p = 3/4. The number of trials is βnβ = 3, and the number of successes we want is βkβ = 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 βn choose kβ. The binomial coefficient represents the number of ways to choose βkβ items from a set of βnβ items, and is given by:

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

Plugging in the values for βnβ, βkβ, βpβ, and βqβ, 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.

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