content attached with this.

' ' '

' ' .

Given : P(D) 0.015 P(D') 1 P(D) 1 0.015 0.985

i.e. P(D') 0.985

' '

'negative'

p n

Let D and D denote the events that represents disease and

non disease respectively

Let T and T denote the events that represent positive and

test r

−

=  = − = − =

=

.

: ( / ) 0.9 ( / ) 1 ( / ) 1 0.9 0.1

. . ( / ) 0.1

: ( / ') 0.05 ( / ') 1 ( / ') 1 0.05 0.95

. . ( / ') 0.95

(

p n p

n

p n p

n

esults respectively

Given P T D

in

P T D P T D

i e P

dividual actually has the disease given that

T D

Given P T D P T D P T D

i

th

e P T

e tes

D

P

=  = − = − =

=

=  = − = − =

=



)

( / )

( ) ( / D) ( ' )

( ) ( / D) ( ') ( / D')

(0.015)(0.1)

(0.015)(0.1) (0.985)(0.95)

0.0016

. .

n

n

n n

t

indicates that a person does not have the disease

P D T

P D P T by Baye s theorem

P D P T P D P T

individual actually has the disease i e P

=

= +

= +

=

0.0016

given that the test

indicates that a person does not have the disease

  = 

 