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
A box contains 5 coins: 4 regular coins and 1 fake coin. When a regular coin is tossed, the probability 𝑃(ℎ𝑒𝑎𝑑) = 0.5 and for a fake coin, 𝑃(ℎ𝑒𝑎𝑑) = 1. You pick a coin at random and toss it twice, and get two heads. The probability that the coin you have chosen is the fake coin is _______. (rounded off to two decimal places)
- Suppose an unbiased coin is tossed 6 times. Each coin toss is independent of all previous coin tosses. Let $E_1$ be the event that among the second, fourth, and sixth coin tosses, there are at least two heads. Let $E_2$ be the event that among the first, second, third, and fifth coin tosses, there are equal number of heads and tails.
The conditional probability P($E_1$ | $E_2$) is equal to ____________. (rounded off to one decimal place) - A box contains three apples and two oranges. Two fruits are removed randomly in succession. The probability that the first is an apple and the second is an orange is _____ .
- An experiment is performed by rolling an unbiased die once. Events A and B are defined as follows:
Event A: The outcome of the rolled die is a prime number.
Event B: The outcome of the rolled die is an odd number less than 4.
The conditional probability of A given B, denoted by P(A|B), is ______ (rounded off to one decimal place). - Bag I contains 4 white and 6 black balls.
Bag II contains 4 white and 3 black balls.
One ball is drawn at random from any one of the two bags and it is found to be a black ball. The probability that the black ball was drawn from Bag I is ______ (rounded off to two decimal places).
