A biologist uses a common diagnostic test to determine whether plants have a certain disease. The two-way frequency table shows the results after testing 1,000 plants.

\begin{tabular}{l|ccc|}
\hline & Test is positive & Test is negative & Total \\
\hline Has disease & 50 & 0 & 50 \\
Doesn't have disease & 7 & 943 & 950 \\
Total & 57 & 943 & 1,000 \\
\hline
\end{tabular}

Select the correct answer from each drop-down menu.

[tex]$P$[/tex](has disease | test is negative) = [tex]$\square$[/tex] %

[tex]$P$[/tex](false positive | doesn't have disease) = [tex]$\square$[/tex] %

[tex]$P$[/tex](doesn't have disease | test is positive) [tex]$\approx$[/tex] [tex]$\square$[/tex] %



Answer :

To tackle this problem, we need to determine three conditional probabilities based on the given two-way frequency table:

1. [tex]\( P(\text{Has disease} \mid \text{Test is negative}) \)[/tex]
2. [tex]\( P(\text{False positive} \mid \text{Doesn't have disease}) \)[/tex]
3. [tex]\( P(\text{Doesn't have disease} \mid \text{Test is positive}) \)[/tex]

We will analyze each probability step-by-step.

### 1. [tex]\( P(\text{Has disease} \mid \text{Test is negative}) \)[/tex]

This is the probability that a plant has the disease given that the test result is negative. Mathematically, it is expressed as:
[tex]\[ P(\text{Has disease} \mid \text{Test is negative}) = \frac{P(\text{Has disease and Test is negative})}{P(\text{Test is negative})} \][/tex]

From the table:
- The number of plants that have the disease and tested negative is [tex]\(0\)[/tex].
- The total number of plants that tested negative is [tex]\(943\)[/tex].

Thus,
[tex]\[ P(\text{Has disease} \mid \text{Test is negative}) = \frac{0}{943} = 0 \% \][/tex]

### 2. [tex]\( P(\text{False positive} \mid \text{Doesn't have disease}) \)[/tex]

This is the probability that the test gives a false positive result given that the plant doesn't have the disease. A false positive occurs when the test is positive but the plant doesn't have the disease. Mathematically, it is expressed as:
[tex]\[ P(\text{False positive} \mid \text{Doesn't have disease}) = \frac{P(\text{Doesn't have disease and Test is positive})}{P(\text{Doesn't have disease})} \][/tex]

From the table:
- The number of plants that don't have the disease and tested positive is [tex]\(7\)[/tex].
- The total number of plants that don't have the disease is [tex]\(950\)[/tex].

Thus,
[tex]\[ P(\text{False positive} \mid \text{Doesn't have disease}) = \frac{7}{950} \approx 0.7368 \% \][/tex]

### 3. [tex]\( P(\text{Doesn't have disease} \mid \text{Test is positive}) \)[/tex]

This is the probability that a plant doesn't have the disease given that the test result is positive. Mathematically, it is expressed as:
[tex]\[ P(\text{Doesn't have disease} \mid \text{Test is positive}) = \frac{P(\text{Doesn't have disease and Test is positive})}{P(\text{Test is positive})} \][/tex]

From the table:
- The number of plants that don't have the disease and tested positive is [tex]\(7\)[/tex].
- The total number of plants that tested positive is [tex]\(57\)[/tex].

Thus,
[tex]\[ P(\text{Doesn't have disease} \mid \text{Test is positive}) = \frac{7}{57} \approx 12.28 \% \][/tex]

### Summary
Based on the calculations:
- [tex]\( P(\text{Has disease} \mid \text{Test is negative}) = 0\% \)[/tex]
- [tex]\( P(\text{False positive} \mid \text{Doesn't have disease}) \approx 0.7368\% \)[/tex]
- [tex]\( P(\text{Doesn't have disease} \mid \text{Test is positive}) \approx 12.28\% \)[/tex]

Therefore, these are the correct values for the probabilities.