To solve this problem, we will use Bayes’ theorem, which states that:
$$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$
In this case, let’s define the events:
Event A: The person has the disease
Event A′: The person does not have the disease (complementary event)
Event B: The person tests positive
We want to find the probability P(A|B), which represents the probability that a person has the disease given that they tested positive. Now, let’s plug in the given probabilities:
1. P(A): The probability that a person has the disease is given as 1 in 10,000, i.e., $P(A) = \frac{1}{10,000} = 0.0001$.
2. P(A′): The probability of not having the disease is P(A′) = 1 − P(A) = 0.9999.
3. P(B|A): The probability that a person tests positive given that they have the disease (a true positive) is the accuracy of the test, i.e., P(B|A) = 0.99.
4. P(B|A′): The probability that a person tests positive given that they do not have the disease (a false positive) is the complement of the test’s accuracy, i.e., P(B|A′) = 1 − P(B|A) = 0.01.
To find P(B) (the probability of testing positive), we use the law of total probability:
P(B) = P(B|A)P(A) + P(B|A′)P(A′)
Now, we can plug in the values we have:
P(B) = 0.99 × 0.0001 + 0.01 × 0.9999 ≈ 0.000099 + 0.009999 ≈ 0.010098
With all the necessary probabilities, we can now apply Bayes’ theorem to find P(A|B):
$$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{0.99 \times 0.0001}{0.010098} \approx 0.0098$$
Therefore, the probability that a person actually has the disease given that they tested positive is about 0.98% (or 0.0098 in decimal form).