In a study 42% of adults questioned reported that their health was excellent. A researcher wishes to study the health 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.

To find the probability that when 13 adults are randomly selected, 3 or fewer are in excellent health, we can use the binomial probability formula.

The binomial probability formula is:
P(X = k) = (n C k) * p^k * (1-p)^(n-k)

Where:
P(X = k) is the probability of having exactly k successes
n is the number of trials
k is the number of successes
p is the probability of success in one trial
(1-p) is the probability of failure in one trial
(n C k) is the number of combinations of n items taken k at a time

In this case, n = 13 (number of adults selected), k can be 3 or fewer (0, 1, 2, 3), p = 0.42 (probability of excellent health based on the study), and (1-p) = 1 - 0.42 = 0.58.

So we need to calculate the sum of probabilities for each value of k from 0 to 3 as follows:
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Using the binomial probability formula, the calculations are as follows:

P(X = 0) = (13 C 0) * 0.42^0 * 0.58^13
P(X = 1) = (13 C 1) * 0.42^1 * 0.58^12
P(X = 2) = (13 C 2) * 0.42^2 * 0.58^11
P(X = 3) = (13 C 3) * 0.42^3 * 0.58^10

To calculate each combination (n C k), we can use the formula:
(n C k) = n! / (k! * (n-k)!)

After calculating each probability, we can simply add them up to get the final probability.

(Note: The exclamation mark "!" denotes factorial, which means multiplying all positive integers up to that number. For example, 4! = 4 * 3 * 2 * 1 = 24)

Let's calculate the probability step by step:

P(X = 0) = (13 C 0) * 0.42^0 * 0.58^13
= 1 * 1 * 0.58^13
≈ 0.003

P(X = 1) = (13 C 1) * 0.42^1 * 0.58^12
= (13! / (1! * (13-1)!)) * 0.42 * 0.58^12
≈ 0.026

P(X = 2) = (13 C 2) * 0.42^2 * 0.58^11
= (13! / (2! * (13-2)!)) * 0.42^2 * 0.58^11
≈ 0.103

P(X = 3) = (13 C 3) * 0.42^3 * 0.58^10
= (13! / (3! * (13-3)!)) * 0.42^3 * 0.58^10
≈ 0.213

Finally, we can add up these probabilities:

P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
≈ 0.003 + 0.026 + 0.103 + 0.213
≈ 0.345

Therefore, the probability that when 13 adults are randomly selected, 3 or fewer are in excellent health is approximately 0.345.

To find the probability that when 13 adults are randomly selected, 3 or fewer are in excellent health, we can use the binomial probability formula.

The binomial probability formula is:

P(x) = (nCx) * p^x * (1-p)^(n-x)

Where:
P(x) represents the probability of getting exactly x successes
n is the number of trials
p is the probability of success in a single trial
(nCx) represents the number of combinations

In this case, we want to find the probability of getting 3 or fewer adults in excellent health out of 13 adults. The probability of an adult being in excellent health is 42% or 0.42.

Using the binomial probability formula, the probability can be calculated as follows:

P(0 or 1 or 2 or 3) = P(0) + P(1) + P(2) + P(3)

P(0) = (13C0) * 0.42^0 * (1-0.42)^(13-0)
P(1) = (13C1) * 0.42^1 * (1-0.42)^(13-1)
P(2) = (13C2) * 0.42^2 * (1-0.42)^(13-2)
P(3) = (13C3) * 0.42^3 * (1-0.42)^(13-3)

To calculate these probabilities, we need to evaluate the combinations and perform the calculations.

(13C0) = 1
(13C1) = 13
(13C2) = (13!)/(2!(13-2)!) = 78
(13C3) = (13!)/(3!(13-3)!) = 286

Now let's substitute these values into the formula and calculate the probabilities:

P(0) = 1 * 0.42^0 * 0.58^13 = 0.007
P(1) = 13 * 0.42^1 * 0.58^12 = 0.040
P(2) = 78 * 0.42^2 * 0.58^11 = 0.118
P(3) = 286 * 0.42^3 * 0.58^10 = 0.228

Finally, sum up these probabilities to find the probability of getting 3 or fewer adults in excellent health:

P(3 or fewer) = 0.007 + 0.040 + 0.118 + 0.228 = 0.393

Therefore, the probability that when 13 adults are randomly selected, 3 or fewer are in excellent health is approximately 0.393.

0.096