In: Statistics and Probability
In a certain risk group, individuals are tested for a certain disease S. A person who has the disease gets the correct diagnosis with probability 0.99, whereas a person who does not suffer from S gets the correct diagnosis with probability 0.95. Furthermore, it is known that 6% of the individuals in the group get the diagnosis “suffer from S”. Determine a) the proportion of individuals in the group who suffer from S, and b) the probability that a person who gets the diagnosis “suffer from S” carries the disease.
Answer:-
Given That:-
In a certain risk group, individuals are tested for a certain disease S. A person who has the disease gets the correct diagnosis with probability 0.99, whereas a person who does not suffer from S gets the correct diagnosis with probability 0.95. Furthermore, it is known that 6% of the individuals in the group get the diagnosis “suffer from S”. Determine a) the proportion of individuals in the group who suffer from S, and b) the probability that a person who gets the diagnosis “suffer from S” carries the disease.
(1) The probability that a person gets the diagnosis "Suffer from fever " actually carries the disease is 0.95
(2) The probability that a randomly selected person has fever or correct doagnosis is 0.061
Step-by - step explanation:
Denote the events as follows:
X = a person carries the disease
Y=the diagnosis is correct .
The information probvided is :
(1) The conditional probability of an event B given that another event A has already occurred is:
Use the conditional probability rule to determine the value of the asked probability
The probability that a person gets the diagnosis "suffer from fever " actual carries the disease can be expressed as follows :
The law of total probability states that:
Compute the probability of event X using the law of total probability as follows:
The value of P(X\Y) is:
Thus, the probability that a person gets the diagnosis "Suffer from fever" actually carries the disease is 0.96