Question

# The chance that Doctor A will diagonise disease X correctly is $$60\%$$. The chance that a patient will die by his treatment after correct diagnosis is $$40\%$$ and the chance of death after wrong diagnosis is $$70\%.$$ A patient of Doctor A who had disease X died. The probability that his disease was diagonised correctly is

A
513
B
613
C
213
D
713

Solution

## The correct option is D $$\dfrac{6}{13}$$Let us define the following events.$${ E }_{ 1 }:$$ Disease $$X$$ is diagnosed correctly by doctor $$A$$.$${ E }_{ 2 }:$$ Disease $$X$$ is not diagnosed correctly by doctor $$A$$.$$B:$$ A patient (of doctor $$A$$) who has disease $$X$$ dies.Then, we are given, $$P({ E }_{ 1 })=0.6$$,$$P({ E }_{ 2 })=1-P({ E }_{ 1 })=1-0.6=0.4$$and $$\displaystyle P\left( \frac { B }{ { E }_{ 1 } } \right) =0.4$$, $$\displaystyle P\left( \frac { B }{ { E }_{ 2 } } \right) =0.7$$By Bay's Theorem$$\displaystyle P\left( \frac { { E }_{ 1 } }{ B } \right) =\frac { P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } } \right) }{ P\left( { E }_{ 1 } \right) P\left( \frac { B }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) P\left( \frac { B }{ { E }_{ 2 } } \right) }$$$$\displaystyle=P\left( \frac { { E }_{ 1 } }{ B } \right) =\frac { 0.6\times 0.4 }{ 0.6\times 0.4+0.4\times 0.7 } =\frac { 0.24 }{ 0.24+0.28 } =\frac { 0.24 }{ 0.52 } =\frac { 6 }{ 13 }$$Mathematics

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