Answer :
To solve the problem, we can use concepts from probability, specifically involving the geometric distribution, which deals with the number of Bernoulli trials needed to get the first success.
Number of People Expected to Check:
Given that 10% of American adults are allergic to dust mites, the probability [tex]p[/tex] of a randomly chosen person being allergic is 0.1. The geometric distribution can describe the scenario where we want to find the number of trials until the first success (a person allergic to dust mites). The expected number of trials for a geometric distribution is given by [tex]\frac{1}{p}[/tex].
Therefore, the researcher should expect to check:
[tex]\text{Expected number of people} = \frac{1}{0.1} = 10[/tex]
This means they should check approximately 10 people to find one who is allergic.
Probability that the First Allergic Person is the Sixth Person Checked:
For a geometric distribution, the probability [tex]P(X = k)[/tex] that the first success occurs on the [tex]k^{th}[/tex] trial is given by:
[tex]P(X = k) = (1-p)^{k-1} \cdot p[/tex]
Here, [tex]p = 0.1[/tex] and [tex]k = 6[/tex]:
[tex]P(X = 6) = (1-0.1)^{6-1} \cdot 0.1 = (0.9)^5 \cdot 0.1[/tex]
[tex]P(X = 6) = 0.59049 \cdot 0.1 = 0.059049[/tex]
Thus, the probability that the first person who is allergic is the sixth person checked is approximately 0.059.
Probability of Finding Someone Allergic Before the Fourth Person:
Method 1: Cumulative Probability
We need the cumulative probability of finding someone allergic within the first three checks (i.e., before the fourth).
[tex]P(X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3)[/tex]
Using the formula for the geometric distribution:
[tex]P(X = 1) = (0.9)^{0} \cdot 0.1 = 0.1[/tex]
[tex]P(X = 2) = (0.9)^{1} \cdot 0.1 = 0.09[/tex]
[tex]P(X = 3) = (0.9)^{2} \cdot 0.1 = 0.081[/tex][tex]P(X \leq 3) = 0.1 + 0.09 + 0.081 = 0.271[/tex]
Method 2: Complement Probability
Alternatively, calculate the probability of not finding an allergic person in the first three checks and subtract from 1.
[tex]P(X > 3) = (1-0.1)^3 = (0.9)^3 = 0.729[/tex]
[tex]P(X \leq 3) = 1 - P(X > 3) = 1 - 0.729 = 0.271[/tex]
In both methods, the probability that the researcher finds someone who is allergic to dust mites before checking the fourth person is 0.271.