Suppose that during a pandemic, 2 percent of people in a region test positive for a virus. You repeatedly test people from this region until you find someone who tests positive, and you are interested in the probability that the first positive test will be the 11th person tested.

Because we keep testing until the first positive test, this situation can be modeled using a Geometric Distribution with:

a. Success [tex]$=$[/tex]
- A person tests positive
- A person does not test positive
- A person gets tested
- 11 people test positive

b. [tex]$p=$[/tex] [tex]$0.02$[/tex]

c. [tex]$X=$[/tex] [tex]$\square$[/tex]

Enter an integer or decimal number.



Answer :

To solve this problem, we utilize the properties of the Geometric Distribution, which is appropriate for modeling the number of trials needed to get the first success in a sequence of independent and identically distributed Bernoulli trials.

Let's break down the question step by step:

### a. Defining the Success Event
- Success is defined as the event where a person tests positive for the virus.

### b. Probability of Success ([tex]\( p \)[/tex])
- Given that 2 percent of people test positive for the virus, the probability [tex]\( p \)[/tex] of success (i.e., testing positive) is 0.02.

### c. Defining [tex]\( X \)[/tex]
- [tex]\( X \)[/tex] represents the number of tests needed to get the first positive result.
- We are specifically interested in the probability that the first positive test occurs on the 11th test.

### Applying the Geometric Distribution
We use the formula for the Geometric Distribution, which calculates the probability of having the first success on the [tex]\( k \)[/tex]-th trial:

[tex]\[ P(X = k) = (1-p)^{k-1} \cdot p \][/tex]

By substituting the known values into the formula:
- [tex]\( p = 0.02 \)[/tex]
- [tex]\( k = 11 \)[/tex]
- [tex]\( (1 - p) = 1 - 0.02 = 0.98 \)[/tex]

We calculate the probability as follows:
[tex]\[ P(X = 11) = (0.98)^{10} \cdot 0.02 \][/tex]

This means that we raise 0.98 to the power of 10 (since [tex]\( k-1 = 11-1 = 10 \)[/tex]), and then multiply the result by 0.02.

### Result
The calculated probability of the first positive test occurring on the 11th test is:
[tex]\[ P(X = 11) = 0.016341456137750933 \][/tex]

Thus, the probability that the first positive test will be on the 11th person tested is approximately 0.0163 or 1.63%.