Recently the County Labor and public Welfare investigated the feasibility of set

Question

Recently the County Labor and public Welfare investigated the feasibility of setting up a county-wide screening program to detect child abuse. A team of consultants estimated the following probabilities:

1 child in 90 is abused

A physician can detect an abused child 90% of the time.

A screening program would incorrectly label 3% of all non-abused children as abused.

A). What is the probability that a child is actually abused given that the screening program diagnosed this child as abused? (Please construct a table when answering this question.)

B). How does the answer in part A change if the incidence of abuse is in 1000? (Please construct a table when answering this question.)

To whom it may concern,

I would like to thank you in advance for your help.

Explanation / Answer

Solution-

Let the events are-

A- Child is actually abused.

N- Child is not abused

D- Labelled as abused

Given probabilities are-

P(A) = 1/90 and P(N) = 89/90

P(D|A) = 0.9 and P(D|N) = 0.03

(A) P(A|D)

= P(A) * P(D|A) / [ P(A) * P(D|A) + P(N) * P(D|N) ] { using bayes theorem }

= 1/90 * 0.9 / [ 1/90 * 0.9 + 89/90 * 0.03]

= 0.2521

(B) Now, P(A) 1/1000

SO Answer would be-

= P(A) * P(D|A) / [ P(A) * P(D|A) + P(N) * P(D|N) ] { using bayes theorem }

= 1/1000 * 0.9 / [ 1/1000 * 0.9 + 999/1000 * 0.03]

= 0.02916

Answers

TY!

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