Question 1166513: A disease test has a false positive rate of 0.5% and a false negative rate of 12%. Compute the probability that someone who tests negative does not have the virus for the following case:
Using the test on a population, where the probability of someone having the disease is 6/100000.
Answer by CPhill(2030)   (Show Source):
You can put this solution on YOUR website! This is a classic application of **Bayes' Theorem** to find the posterior probability.
The probability that someone who tests negative does not have the virus is approximately **99.9999%** (or **0.999999**).
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## 🔬 Probability Calculation
Let $D$ be the event that a person **Has the Disease**.
Let $D'$ be the event that a person **Does Not Have the Disease**.
Let $T'$ be the event that the test result is **Negative**.
### 1. Identify the Given Probabilities
* **Prevalence ($P(D)$):** The probability of having the disease in the population.
$$P(D) = \frac{6}{100000} = 0.00006$$
* **Probability of Not Having the Disease ($P(D')$):**
$$P(D') = 1 - P(D) = 1 - 0.00006 = 0.99994$$
* **False Positive Rate ($FP$):** The probability of testing positive when the person does *not* have the disease.
$$P(T|D') = 0.5\% = 0.005$$
* **False Negative Rate ($FN$):** The probability of testing negative when the person *does* have the disease.
$$P(T'|D) = 12\% = 0.12$$
### 2. Required Conditional Probabilities
We need the probability of correctly testing negative ($P(T'|D')$):
* **True Negative Rate ($TN$):** The probability of testing negative when the person does *not* have the disease.
$$P(T'|D') = 1 - P(T|D') = 1 - 0.005 = 0.995$$
### 3. Goal
We want to compute the probability that someone who tests negative does not have the disease, $P(D'|T')$.
$$P(D'|T') = \frac{P(T'|D') \cdot P(D')}{P(T')}$$
### 4. Calculate the Total Probability of Testing Negative ($P(T')$)
The Law of Total Probability states that a negative result can occur in two mutually exclusive ways: a true negative ($T'$ and $D'$) or a false negative ($T'$ and $D$).
$$P(T') = P(T'|D') \cdot P(D') + P(T'|D) \cdot P(D)$$
Substitute the values:
$$P(T') = (0.995 \cdot 0.99994) + (0.12 \cdot 0.00006)$$
$$P(T') = 0.9949403 + 0.0000072$$
$$P(T') = 0.9949475$$
### 5. Apply Bayes' Theorem
Now, substitute $P(T')$ back into the formula for $P(D'|T')$:
$$P(D'|T') = \frac{P(T'|D') \cdot P(D')}{P(T')}$$
$$P(D'|T') = \frac{0.995 \cdot 0.99994}{0.9949475}$$
$$P(D'|T') = \frac{0.9949403}{0.9949475}$$
$$P(D'|T') \approx \mathbf{0.9999927}$$
Rounding to six decimal places, the probability is **0.999993**.
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