. In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a disease. The probability that a person carries the gene (independently of other people) is 0.1. What is the probability that we will have to test at least 7 people in order to find 4 people who carry the gene? (b) .00255 (c) .9987 (d) 1.458 x 10-6 (a) .9982 (e) .00346
The Correct Answer and Explanation is:
To solve this problem, we need to understand the Negative Binomial distribution, which models the number of trials needed to achieve a certain number of successes in independent Bernoulli trials.
Given:
- The probability that a person carries the gene (a “success”) is p=0.1p = 0.1.
- We are interested in finding 4 people who carry the gene.
- We want to know the probability that we must test at least 7 people to find these 4 carriers.
Restating the question:
Let XX be the number of people tested to find the 4th carrier.
We want to compute: P(X≥7)P(X \geq 7)
This is the complement of the cumulative probability up to 6 people: P(X≥7)=1−P(X≤6)P(X \geq 7) = 1 – P(X \leq 6)
Here, X∼Negative Binomial(r=4,p=0.1)X \sim \text{Negative Binomial}(r=4, p=0.1), which counts the number of trials until the 4th success.
We now calculate P(X≤6)P(X \leq 6), i.e., the probability that the 4th person who carries the gene is found on or before the 6th person.
Use the Negative Binomial probability mass function: P(X=x)=(x−1r−1)pr(1−p)x−rP(X = x) = \binom{x-1}{r-1} p^r (1-p)^{x – r}
For r=4r = 4, we calculate for x=4,5,6x = 4, 5, 6:
- P(X=4)=(33)(0.1)4(0.9)0=1⋅0.0001=0.0001P(X=4) = \binom{3}{3} (0.1)^4 (0.9)^0 = 1 \cdot 0.0001 = 0.0001
- P(X=5)=(43)(0.1)4(0.9)1=4⋅0.0001⋅0.9=0.00036P(X=5) = \binom{4}{3} (0.1)^4 (0.9)^1 = 4 \cdot 0.0001 \cdot 0.9 = 0.00036
- P(X=6)=(53)(0.1)4(0.9)2=10⋅0.0001⋅0.81=0.00081P(X=6) = \binom{5}{3} (0.1)^4 (0.9)^2 = 10 \cdot 0.0001 \cdot 0.81 = 0.00081
So, P(X≤6)=0.0001+0.00036+0.00081=0.00127P(X \leq 6) = 0.0001 + 0.00036 + 0.00081 = 0.00127 P(X≥7)=1−0.00127=0.99873P(X \geq 7) = 1 – 0.00127 = 0.99873
Correct answer: (c) 0.9987
Explanation
This problem involves using the Negative Binomial distribution, which is appropriate when you want to determine the number of independent trials required to achieve a fixed number of successes. In this scenario, a “success” is defined as finding a person who carries a particular gene, and each person is tested independently with a 10% chance (0.1 probability) of being a carrier.
We are told to find the probability that at least 7 people must be tested to find the 4th carrier of the gene. That is, the first 3 carriers are found somewhere in the first 6 people, but the 4th has not yet appeared by the 6th trial. This means the 4th carrier appears in the 7th trial or later.
The Negative Binomial distribution gives us the probability that the r-th success occurs on the x-th trial. Here, r = 4 (we’re looking for the 4th carrier), and we want the cumulative probability up to x = 6. We compute the probabilities of getting the 4th success on trials 4, 5, and 6 using the negative binomial probability mass function, and subtract their sum from 1 to get the probability of needing at least 7 trials.
After computing these values, we find that the total probability of getting the 4th success within the first 6 trials is approximately 0.00127. Therefore, the chance that we will need to test 7 or more people is: 1−0.00127=0.998731 – 0.00127 = 0.99873
Thus, the correct answer is (c) 0.9987.
