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

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

$$\mathrm{Let}\mathrm{A}\mathrm{denote}\mathrm{the}\mathrm{event}\mathrm{of}{}^{\prime }\mathrm{test}{\mathrm{positive}}^{\prime }$$$$\mathrm{in}\mathrm{a}\mathrm{specific}\mathrm{experiment},\mathrm{and}\mathrm{B}$$$$\mathrm{that}\mathrm{of}{}^{\prime }\mathrm{have}{\mathrm{cancer}}^{\prime }.\mathrm{It}\mathrm{is}\mathrm{known}\mathrm{that}$$$$\mathrm{P}\left(\mathrm{B}\right)=0.005,\mathrm{P}(\mathrm{A}\mid \mathrm{B})=0.95\mathrm{and}\mathrm{P}(\stackrel{-}{\mathrm{A}}\mid \stackrel{-}{\mathrm{B}})=0.95$$$$\mathrm{Suppose}\mathrm{someone}\mathrm{tests}\mathrm{positive}\mathrm{in}$$$$\mathrm{the}\mathrm{experiment}.\mathrm{Find}\mathrm{the}\mathrm{probability}\mathrm{that}$$$$\mathrm{he}\mathrm{has}\mathrm{cancer}.$$

Answered by EDWIN88 last updated on 22/Feb/21

$$\mathrm{P}(\mathrm{B}\mid \mathrm{A})=\frac{\mathrm{P}\left(\mathrm{B}\right)\times \mathrm{P}(\mathrm{A}\mid \mathrm{B})}{\mathrm{P}\left(\mathrm{A}\right)}$$$$=\frac{0.005\times 0.95}{0.005\times 0.95+0.995\times 0.05}$$$$=\frac{19}{218}$$

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