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Sagot :
Answer:
0.2916 = 29.16% probability that one of them is pre-diabetic
Step-by-step explanation:
For each adult, there are only two possible outcomes. Either they have blood chemistry parameters consistent with a diagnosis of a pre-diabetic condition, or they do not. Each volunteer in independent of other volunteers. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
10% of the adult population has blood chemistry parameters consistent with a diagnosis of a pre-diabetic condition.
This means that [tex]p = 0.1[/tex]
Of four volunteer participants in a health screening study, what is the probability that one of them is pre-diabetic
This is P(X = 1) when n = 4. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{4,1}.(0.1)^{1}.(0.9)^{3} = 0.2916[/tex]
0.2916 = 29.16% probability that one of them is pre-diabetic
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