Solveeit Logo

Question

Question: The chance that Doctor \[A\] will diagnose disease \[X\] correctly is 60%. The chance that a patient...

The chance that Doctor AA will diagnose disease XX correctly is 60%. The chance that a patient will die by his treatment after the correct diagnosis is 40% and the chance of death after the wrong diagnosis is 70%. A patient of Doctor AA who had disease XX died. The probability that his disease was diagnosed correctly is
A. 513\dfrac{5}{{13}}
B. 613\dfrac{6}{{13}}
C. 213\dfrac{2}{{13}}
D. 713\dfrac{7}{{13}}

Explanation

Solution

In order to find the required probability first we will be given events and conditions as a variable and we will proceed further by using Bayes’ theorem which describes the probability of an event based on prior knowledge of conditions that might be related to the event.

Complete step-by-step solution:
Let us define the following events as:
E1{E_1}: Disease is diagnosed correctly by doctor AA.
E2{E_2}: Disease is not diagnosed correctly by doctor AA.
BB: A patient (of doctor AA) who has disease XX dies.
Given that, P(E1)=60100=0.6P\left( {{E_1}} \right) = \dfrac{{60}}{{100}} = 0.6
We know that, P(E1)+P(E2)=1P\left( {{E_1}} \right) + P\left( {{E_2}} \right) = 1
So, we have P(E2)=1P(E1)=10.6=0.4P\left( {{E_2}} \right) = 1 - P\left( {{E_1}} \right) = 1 - 0.6 = 0.4
Also given that, the probability of BBgiven when E1{E_1} is true is P(BE1)=40100=0.4P\left( {\dfrac{B}{{{E_1}}}} \right) = \dfrac{{40}}{{100}} = 0.4
Probability of BBgiven when E2{E_2} is true is P(BE2)=70100=0.7P\left( {\dfrac{B}{{{E_2}}}} \right) = \dfrac{{70}}{{100}} = 0.7
Now, we have to find the probability of E1{E_1} when BB is true i.e., P(E1B)P\left( {\dfrac{{{E_1}}}{B}} \right)
By Bayes’ Theorem, we have

P(E1B)=P(E1)P(BE1)P(E1)P(BE1)+P(E2)P(BE2) P(E1B)=0.6×0.40.6×0.4+0.4×0.7 P(E1B)=0.240.24+0.28 P(E1B)=0.240.52=613  \Rightarrow P\left( {\dfrac{{{E_1}}}{B}} \right) = \dfrac{{P\left( {{E_1}} \right)P\left( {\dfrac{B}{{{E_1}}}} \right)}}{{\,P\left( {{E_1}} \right)P\left( {\dfrac{B}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\dfrac{B}{{{E_2}}}} \right)}} \\\ \Rightarrow P\left( {\dfrac{{{E_1}}}{B}} \right) = \dfrac{{0.6 \times 0.4}}{{0.6 \times 0.4 + 0.4 \times 0.7}} \\\ \Rightarrow P\left( {\dfrac{{{E_1}}}{B}} \right) = \dfrac{{0.24}}{{0.24 + 0.28}} \\\ \therefore P\left( {\dfrac{{{E_1}}}{B}} \right) = \dfrac{{0.24}}{{0.52}} = \dfrac{6}{{13}} \\\

Therefore, the correct option is B.

Note: Bayes' theorem is a mathematical formula for determining conditional probability. Conditional probability is the likelihood of an outcome occurring, based on a previous outcome occurring. Bayes' theorem provides a way to revise existing predictions or theories (update probabilities) given new or additional evidence.