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Question Number 133420 by liberty last updated on 22/Feb/21

Let A denote the event of ′test positive′  in a specific experiment, and B  that of ′have cancer′. It is known that  P(B)=0.005 , P(A∣B)= 0.95 and P(A^− ∣B^− )= 0.95  Suppose someone tests positive in  the experiment. Find the probability that  he has cancer .

$$\mathrm{Let}\:\mathrm{A}\:\mathrm{denote}\:\mathrm{the}\:\mathrm{event}\:\mathrm{of}\:'\mathrm{test}\:\mathrm{positive}' \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{specific}\:\mathrm{experiment},\:\mathrm{and}\:\mathrm{B} \\ $$$$\mathrm{that}\:\mathrm{of}\:'\mathrm{have}\:\mathrm{cancer}'.\:\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that} \\ $$$$\mathrm{P}\left(\mathrm{B}\right)=\mathrm{0}.\mathrm{005}\:,\:\mathrm{P}\left(\mathrm{A}\mid\mathrm{B}\right)=\:\mathrm{0}.\mathrm{95}\:\mathrm{and}\:\mathrm{P}\left(\overset{−} {\mathrm{A}}\mid\overset{−} {\mathrm{B}}\right)=\:\mathrm{0}.\mathrm{95} \\ $$$$\mathrm{Suppose}\:\mathrm{someone}\:\mathrm{tests}\:\mathrm{positive}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{experiment}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\mathrm{he}\:\mathrm{has}\:\mathrm{cancer}\:. \\ $$

Answered by EDWIN88 last updated on 22/Feb/21

P(B∣A) = ((P(B)×P(A∣B))/(P(A)))                   = ((0.005×0.95)/(0.005×0.95+0.995×0.05))                  = ((19)/(218))

$$\mathrm{P}\left(\mathrm{B}\mid\mathrm{A}\right)\:=\:\frac{\mathrm{P}\left(\mathrm{B}\right)×\mathrm{P}\left(\mathrm{A}\mid\mathrm{B}\right)}{\mathrm{P}\left(\mathrm{A}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{0}.\mathrm{005}×\mathrm{0}.\mathrm{95}}{\mathrm{0}.\mathrm{005}×\mathrm{0}.\mathrm{95}+\mathrm{0}.\mathrm{995}×\mathrm{0}.\mathrm{05}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{19}}{\mathrm{218}} \\ $$

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