Math  /  Data & Statistics

QuestionIn 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 is 0.1 What is the probability FOUR OR MORE people will have to be tested before TWO with the gene are detected? (Round to the nearest two decimals 0.00)
Answer:

Studdy Solution

STEP 1

What is this asking? We need to find the probability that we test **four or more people** before finding **two people** with the gene. Watch out! Don't confuse the probability of finding one person with the gene with finding two people.
Also, remember that "four or more" includes four, five, six, and so on!

STEP 2

1. Define the problem as a negative binomial distribution
2. Calculate the probability for exactly three tests
3. Calculate the probability for four or more tests
4. Sum the probabilities

STEP 3

Alright, let's start by understanding that this is a **negative binomial distribution** problem.
Why? Because we're looking for the probability of a certain number of failures before achieving a fixed number of successes.
Here, a "success" is finding someone with the gene, and a "failure" is testing someone without it.

STEP 4

The probability of finding someone with the gene is given as p=0.1 p = 0.1 .
So, the probability of not finding someone with the gene is 1p=0.9 1 - p = 0.9 .

STEP 5

To find the probability that exactly three people are tested before finding two with the gene, we need to have one failure and two successes in three tests.
The formula for the negative binomial distribution is:
P(X=k)=(k1r1)pr(1p)krP(X = k) = \binom{k-1}{r-1} \cdot p^r \cdot (1-p)^{k-r}where k k is the total number of trials, r r is the number of successes, and p p is the probability of success.

STEP 6

Plug in the values: k=3 k = 3 , r=2 r = 2 , and p=0.1 p = 0.1 :
P(X=3)=(3121)(0.1)2(0.9)32P(X = 3) = \binom{3-1}{2-1} \cdot (0.1)^2 \cdot (0.9)^{3-2}

STEP 7

Calculate the binomial coefficient and the probabilities:
P(X=3)=(21)0.010.9=20.010.9=0.018P(X = 3) = \binom{2}{1} \cdot 0.01 \cdot 0.9 = 2 \cdot 0.01 \cdot 0.9 = 0.018

STEP 8

Now, we need the probability of getting two successes in four or more tests.
This means we need to find the probability of **not** getting two successes in the first three tests.

STEP 9

The probability of not getting two successes in the first three tests is the complement of getting two successes in exactly three tests:
P(X4)=1P(X=3)P(X \geq 4) = 1 - P(X = 3)

STEP 10

Substitute the value we found earlier:
P(X4)=10.018=0.982P(X \geq 4) = 1 - 0.018 = 0.982

STEP 11

The probability that four or more people will have to be tested before finding two with the gene is **0.98**.

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