89. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances 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 and yoga.
89. Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances 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 and yoga.
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1 Answer
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89. A patient has options to have the treatment of yoga and meditation and that of prescription of drugs.
Let these events be denoted by E1 and E2 i.e.,
E1 = Treatment of yoga and meditation
E2 = Treatment of prescription of certain drugs
P (E1) = P (E2) = 1/2
Let A denotes that a person has heart attack, then P (A) = 40% = 0.40
Yoga and meditation reduces heart attack by 30.
Inspite of getting yoga and meditation treatment heart risk is 70% of 0.40
P (A|E1) = 0.40 x 0.70 = 0.28
Also, Drug prescription reduces the heart attack rick by 25%
Even after adopting the drug prescription hear rick is 75% of 0.40
P (A|E2)&
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P (2 obtained on even numbered toss) = k (let)
P (2) =
P (
If x = 0, y = 6, 7, 8, 9, 10
If x = 1, y = 7, 8, 9, 10
If x = 2, y = 8, 9, 10
If x = 3, y = 9, 10
If x = 4, y = 10
If x = 5, y = no possible value
Total possible ways = (5 + 4 + 3 + 2 + 1) * 2
= 30
Required probability
P (2W and 2B) = P (2B, 6W) × P (2W and 2B)
+ P (3B, 5W) × P (2W and 2B)
+ P (4B, 4W) × P (2W and 2B)
+ P (5B, 3W) × P (2W and 2B)
+ P (6B, 2W) × P (2W and 2B)
(15 + 30 + 36 + 30 + 15)
Let probability of tail is
⇒ Probability of getting head =
∴ Probability of getting 2 heads and 1 tail
ax2 + bx + c = 0
D = b2 – 4ac
D = 0
b2 – 4ac = 0
b2 = 4ac
(i) AC = 1, b = 2 (1, 2, 1) is one way
(ii) AC = 4, b = 4
(iii) AC = 9, b = 6, a = 3, c = 3 is one way
1 + 3 + 1 = 5 way
Required probability =
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