A study revealed that if a person suffers from the disease D, the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease D, the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease D occurs in 30% of the individuals.
If a person tests positive for D in that clinic, the probability that he/she actually suffers from the disease D is __________ . (Rounded off to two decimal places)
Disease D Positive Test Probability Calculation
This problem requires calculating the probability of having disease D given a positive test result, utilizing conditional probabilities and Bayes' Theorem.
Given Probabilities for Disease D Test
- Probability of testing positive if the person has disease D (Sensitivity): $P(\text{Positive}|\text{D}) = 0.80$
- Probability of testing negative if the person has disease D: $P(\text{Negative}|\text{D}) = 0.20$
- Probability of testing positive if the person does not have disease D (False Positive Rate): $P(\text{Positive}|\text{Not D}) = 0.10$
- Probability of testing negative if the person does not have disease D: $P(\text{Negative}|\text{Not D}) = 0.90$
- Prevalence of disease D in the general population: $P(\text{D}) = 0.30$
From the prevalence, we can deduce the probability of not having disease D:
$P(\text{Not D}) = 1 - P(\text{D}) = 1 - 0.30 = 0.70$Calculating Total Probability of a Positive Test
To find the probability of having disease D given a positive test ($P(\text{D}|\text{Positive})$), we first need the overall probability of testing positive ($P(\text{Positive})$).
Using the law of total probability:
$ P(\text{Positive}) = P(\text{Positive}|\text{D}) \times P(\text{D}) + P(\text{Positive}|\text{Not D}) \times P(\text{Not D}) $Substitute the given values:
$ P(\text{Positive}) = (0.80 \times 0.30) + (0.10 \times 0.70) $ $ P(\text{Positive}) = 0.24 + 0.07 $ $ P(\text{Positive}) = 0.31 $Applying Bayes' Theorem for Disease D Probability
Now, apply Bayes' Theorem to find the probability of having disease D given a positive test result:
$ P(\text{D}|\text{Positive}) = \frac{P(\text{Positive}|\text{D}) \times P(\text{D})}{P(\text{Positive})} $Substitute the calculated and given values:
$ P(\text{D}|\text{Positive}) = \frac{0.80 \times 0.30}{0.31} $ $ P(\text{D}|\text{Positive}) = \frac{0.24}{0.31} $Final Probability Calculation and Rounding
Calculating the final value:
$ P(\text{D}|\text{Positive}) \approx 0.77419... $Rounding to two decimal places, the probability is $0.77$. This means that if a person tests positive for disease D, there is approximately a $77\%$ chance they actually have the disease.
Important Questions from Conditional Probability
Two events A and B are such that P(not B) = 0.8, P(A ∪ B) = 0.5 and P(A|B) = 0.4. Then P(A) is equal to
For two events, A and B, it is given that \({\rm{P}}\left( {\rm{A}} \right) = \frac{3}{5},{\rm{\;P}}\left( {\rm{B}} \right) = \frac{3}{{10}}\) and \({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{2}{3}\) . If A̅ and B̅ are the complementary events of A and B, then what is P(A̅ | B̅) equal to?
For two mutually exclusive events A and B, P(A) = 0.2 and P (A̅ ∩ B) = 0.3. What is P (A|(A ∪ B)) equal to?
If an event B has occurred and has P(B) = 1, the conditional probability P(A|B) is equal to:
If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
