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A certain virus infects one in every 200 people. A test used to detect the virus in a person is positive [tex]$85 \%$[/tex] of the time if the person has the virus and positive [tex]$8 \%$[/tex] of the time if the person does not have the virus. (This [tex]$8 \%$[/tex] result is called a false positive.) Let [tex]$A$[/tex] be the event "the person is infected" and [tex]$B$[/tex] be the event "the person tests positive."

\begin{tabular}{|c|l|l|l|}
\hline
& Positive & Negative & Totals \\
\hline
Virus & 425 & 75 & 500 \\
\hline
No virus & 7960 & 91540 & 99500 \\
\hline
Totals & 8385 & 91615 & 100000 \\
\hline
\end{tabular}

a. Find the probability that a person has the virus given that they have tested positive, i.e., find [tex]$P(A \mid B)$[/tex]. Round your answer to the nearest hundredth of a percent and do not include a percent sign.
[tex]\[ P(A \mid B) = \square \][/tex]

b. Find the probability that a person does not have the virus given that they test negative, i.e., find [tex]$P(\neg A \mid \neg B)$[/tex]. Round your answer to the nearest hundredth of a percent and do not include a percent sign.
[tex]\[ P(\neg A \mid \neg B) = \square \][/tex]

Sagot :

GinnyAnswer
Certainly! Let's carefully walk through the solution step-by-step using probability concepts.

### Given Information:

1. Positive Cases:
- Virus positive: 425
- No virus positive: 7960

2. Negative Cases:
- Virus negative: 75
- No virus negative: 91540

3. Totals:
- Total positive: 425 (virus positive) + 7960 (no virus positive) = 8385
- Total negative: 75 (virus negative) + 91540 (no virus negative) = 91615
- Total number of people: 100000

### Part (a): Finding [tex]\( P(A \mid B) \)[/tex]

This is the probability that a person has the virus given that they have tested positive.

Using the definition of conditional probability:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]

Where:
- [tex]\( P(A \cap B) \)[/tex] is the probability of testing positive and having the virus.
- [tex]\( P(B) \)[/tex] is the probability of testing positive.

From the data:
- [tex]\( P(A \cap B) = \frac{425}{100000} \)[/tex]
- [tex]\( P(B) = \frac{8385}{100000} \)[/tex]

So,
[tex]\[ P(A \mid B) = \frac{\frac{425}{100000}}{\frac{8385}{100000}} = \frac{425}{8385} \][/tex]

Now, calculate [tex]\(\frac{425}{8385} \)[/tex]:

[tex]\[ P(A \mid B) \approx 0.0507 \][/tex]

Therefore,
[tex]\[ P(A \mid B) = 5.07\% \][/tex]

### Part (b): Finding [tex]\( P(\neg A \mid \neg B) \)[/tex]

This is the probability that a person does not have the virus given that they test negative.

Using the definition of conditional probability:
[tex]\[ P(\neg A \mid \neg B) = \frac{P(\neg A \cap \neg B)}{P(\neg B)} \][/tex]

Where:
- [tex]\( P(\neg A \cap \neg B) \)[/tex] is the probability of testing negative and not having the virus.
- [tex]\( P(\neg B) \)[/tex] is the probability of testing negative.

From the data:
- [tex]\( P(\neg A \cap \neg B) = \frac{91540}{100000} \)[/tex]
- [tex]\( P(\neg B) = \frac{91615}{100000} \)[/tex]

So,
[tex]\[ P(\neg A \mid \neg B) = \frac{\frac{91540}{100000}}{\frac{91615}{100000}} = \frac{91540}{91615} \][/tex]

Now, calculate [tex]\(\frac{91540}{91615} \)[/tex]:

[tex]\[ P(\neg A \mid \neg B) \approx 0.9992 \][/tex]

Therefore,
[tex]\[ P(\neg A \mid \neg B) = 99.92\% \][/tex]

### Final Answers:

a. [tex]\( P(A \mid B) = 5.07 \)[/tex]

b. [tex]\( P(\neg A \mid \neg B) = 99.92 \)[/tex]

These results reflect the respective probabilities rounded to the nearest hundredth of a percent as required.
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