Respuesta :
Answer:
1.98% probability that at least 11 members are categorized as low risk participants in two of three communities
Step-by-step explanation:
To solve this question, we need to use two separate binomial trials.
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.
Probability of at least 11 members being categorized as low risk participants.
12 members in the sample, so [tex]n = 12[/tex]
30% chance to be categorized as high risk participants. So 100-30 = 70% probability of being categorized as low risk participants. So [tex]p = 0.7[/tex]
This probability is
[tex]P(X \leq 11) = P(X = 11) + P(X = 12)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{12,11}.(0.7)^{11}.(0.3)^{1} = 0.0712[/tex]
[tex]P(X = 2) = C_{12,12}.(0.7)^{12}.(0.3)^{0} = 0.0138[/tex]
[tex]P(X \leq 11) = P(X = 11) + P(X = 12) = 0.0712 + 0.0138 = 0.0850[/tex]
What is the probability that at least 11 members are categorized as low risk participants in two of three communities?
For each community, 8.50% probability of at least 11 members being categorized as low risk. So [tex]p = 0.085[/tex]
Three comunities, so [tex]n = 3[/tex]
This probability is P(X = 2).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.085)^{2}.(0.915)^{1} = 0.0198[/tex]
1.98% probability that at least 11 members are categorized as low risk participants in two of three communities
