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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
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B
613
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C
213
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D
713
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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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