In a study, 37 % of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to a nuclear power plant. Among 13 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 13 adults are randomly selected, 3 or fewer are in excellent health. Round to three decimal places.

There are only 2 outcomes of the event i.e. in excellent health or not.
Number of trials is fixed i.e. n = 13
Probability of success is constant
Events are independent i.e. answer of individuals is not influenced by answers of others

This satisfies all the 4 conditions that are needed for a Binomial Experiment. Hence, we can solve the given question using Binomial Probability.

p = 0.37

q = 1 - p = 1 - 0.37 = 0.63

n = 13

We need to find the probability that when 13 adults are randomly selected, 3 or fewer are in excellent health i.e. P( x ≤ 3 )

P( x ≤ 3 ) = P(0) + P(1) + P(2) + P(3)

The formula of Binomial probability is:

[tex]P(x)=^{n}C_{x}(p)^{x}(q)^{(n-x)}[/tex]

So,

[tex]P(0)=^{13}C_{0}(0.37)^{0}(0.63)^{(13-0)}=0.0025\\\\ P(1)=^{13}C_{1}(0.37)^{1}(0.63)^{(13-1)}=0.0188\\\\ P(2)=^{13}C_{2}(0.37)^{2}(0.63)^{(13-2)}=0.0663\\\\ P(3)=^{13}C_{3}(0.37)^{3}(0.63)^{(13-3)}=0.1427[/tex]

Thus,

P( x ≤ 3 ) = P(0) + P(1) + P(2) + P(3) = 0.0025 + 0.0188 + 0.0663 + 0.1427

P( x ≤ 3 ) = 0.2303

Therefore, the probability that when 13 adults are randomly selected, 3 or fewer are in excellent health is 0.230