Question

In: Statistics and Probability

In a certain risk group, individuals are tested for a certain disease S. A person who...

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.

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Expert Solution

Answer:-

Given That:-

(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

orchestra answered 1 year ago

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