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CBSE Class 12 Math 2013 Solved Paper

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Question : 29 of 29
Marks: +1, -0
Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation
Solution:  
Let A, E1E_1, and E2E_2, respectively denote the events that a person has a heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.
∴ P (A) = 0.40
P (E1)(E_1) = P (E2)(E_2) = 12\frac{1}{2}
P (A|E1E_1) = 0.40 × 0.70 = 0.28
P (A|E2E_2) = 0.40 × 0.75 = 0.30
Probability that the patient suffering a heart attack followed a course of meditation and yoga = P (E1E_1|A)
=
P(E1)P(A∣E1)P(E1)P(A∣E1)+P(E2)P(A∣E2)\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}
= 12×0.2812×0.28+12×0.30\frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28 + \frac{1}{2} \times 0.30}
= 2828+30\frac{28}{28+30}
= 2858\frac{28}{58}
= 1429\frac{14}{29}
Now, calculate P (E2E_2|A)
P (E2E_2|A) =
P(E2)P(A∣E2)P(E1)P(A∣E1)+P(E2)P(A∣E2)\frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}
= 12×0.3012×0.28+12×0.30\frac{\frac{1}{2} \times 0.30}{\frac{1}{2} \times 0.28 + \frac{1}{2} \times 0.30}
= 3028+30\frac{30}{28+30}
= 3058\frac{30}{58}
= 1529\frac{15}{29}
Since P (E1E_1|A) < P (E2E_2|A), the course of yoga and meditation is more beneficial for a patient.
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