Question

$Let d_{1}, d_{2}, d_{3}, be three mutually exclusive diseases. Let S = \{S_{1}, S_{2}, S_{3} \dots \dots . S_{6}\} be$ the set of observable symptoms of these diseases. For example $S_{1}$ is the shortness of breath, $S_{2}$ is loss of weight, $S_{3}$ is fatigue etc. suppose a random sample of 10,000 patients contains 3200 patients with disease $d_{1}$ . 3500 with disease $d_{2}$ and 3300 with disease $d_{3}$ . Also, 3100 patients with disease $d_{1}$ , 3300 with disease $d_{2}$ and 3000 with disease $d_{3}$ , show the symptom S. knowing that the patient has symptom S, the doctor wishes to determine the patient’s illness.

Difficult

A

$Let the probability of patient having disease d_{1} be \frac{p}{q} (GCD of (p_{a} g) is 1), then$ the value of p+q is 120
B

$The doctor should conclude that the patient is most likely to have diseases d_{3}$
C

$The doctor should conclude that the patient is most likely to have diseases d_{1}$
D

$The doctor should conclude that the patient is most likely to have diseases d_{2}$

Solution

$Let E_{1} denote the event that the patient has disease d_{j}; i = 1, 2, 3 and A be the$
$\begin{array}{l} P (E_{1}) = \frac{3200}{10000} = \frac{32}{100}, P (E_{2}) = \frac{3500}{10000} = \frac{35}{100} \\ P (E_{3}) = \frac{3300}{10000} = \frac{33}{100} \\ P (A \cap E_{1}) = \frac{3100}{10000} = \frac{31}{100}, P (A \cap E_{2}) = \frac{3300}{10000} = \frac{33}{100} and P (A \cap E_{3}) = \frac{3000}{10000} = \frac{3}{10} \end{array}$
$\begin{aligned} ∴ P (\frac{A}{E_{1}}) = \frac{P (A \cap E_{1})}{P (E_{1})} \\ \Rightarrow P (\frac{A}{E_{1}}) = \frac{\frac{31}{100}}{\frac{32}{100}} = \frac{31}{32} \\ P (\frac{A}{E_{2}}) = \frac{P (A \cap E_{2})}{P (E_{2})} \\ \Rightarrow P (\frac{A}{E_{2}}) = \frac{\frac{33}{100}}{\frac{35}{100}} = \frac{33}{35} \\ P (\frac{A}{E_{3}}) = \frac{P (A \cap E_{3})}{P (E_{3})} \end{aligned}$
$\Rightarrow P (\frac{A}{E_{3}}) = \frac{\frac{3}{10}}{\frac{33}{100}} = \frac{30}{33}$
Using Bayes’ theorem, we have
$\begin{array}{l} P (\frac{E_{1}}{A}) = \frac{P (E_{1}) P (\frac{A}{E_{1}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}}) + P (E_{3}) P (\frac{A}{E_{3}})} \\ = \frac{\frac{32}{100} \times \frac{31}{32}}{\frac{32}{100} \times \frac{31}{32} + \frac{35}{100} \times \frac{33}{35} + \frac{33}{100 \times \frac{30}{33}} = \frac{31}{94}} \\ P (\frac{E_{2}}{A}) = \frac{P (E_{2}) P (\frac{A}{E_{2}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}}) + P (E_{3}) P (\frac{A}{E_{3}})} \end{array}$
$\begin{array}{l} = \frac{\frac{35}{100} \times \frac{33}{35}}{\frac{32}{100} \times \frac{31}{32} + \frac{35}{100} \times \frac{33}{35} + \frac{33}{100} \times \frac{30}{33}} = \frac{33}{94} and \\ P (\frac{E_{3}}{A}) = \frac{P (E_{3}) P (\frac{A}{E_{3}})}{P (E_{1}) P (\frac{A}{E_{1}}) + P (E_{2}) P (\frac{A}{E_{2}}) + P (E_{3}) P (\frac{A}{E_{3}})} \\ = \frac{\frac{33}{100} \times \frac{30}{33}}{\frac{32}{100} \times \frac{31}{32} + \frac{35}{100} \times \frac{33}{35} + \frac{33}{100} \times \frac{30}{33}} = \frac{30}{94} \end{array}$
$P (\frac{E_{3}}{A}) < P (\frac{E_{1}}{A}) < P (\frac{E_{2}}{A})$
is largest. Thus the doctor should conclude that the patient is most likely to have disease $d_{2}$

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