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. $\frac{5}{13}$
• B. $\frac{6}{13}$
• C. $\frac{2}{13}$
• D. $\frac{7}{13}$

Answer 1

Answer: (B)

$\frac{6}{13}$

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, $P\left(E_1\right)=0.6$,$P\left(E_2\right)=1-P\left(E_1\right)=1-0.6=0.4$
and $P\left(\frac{B}{E_1}\right)=0.4$, $P\left(\frac{B}{E_2}\right)=0.7$
By Bay's Theorem
$P\left(\frac{E_1}{B}\right)=\frac{P\left(E_1\right)P\left(\frac{B}{E_1}\right)}{P\left(E_1\right)P\left(\frac{B}{E_1}\right)+P\left(E_2\right)P\left(\frac{B}{E_2}\right)}$
$=P\left(\frac{E_1}{B}\right)=\frac{0.6\times 0.4}{0.6\times 0.4+0.4\times 0.7}=\frac{0.24}{0.24+0.28}=\frac{0.24}{0.52}=\frac{6}{13}$