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Q7:
Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce 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?
Solution :
Step 1: Definition of Events
Let $E$ be the event that the patient suffers a heart attack. Let $A_1$ and $A_2$ be the events of choosing the meditation/yoga course and the drug prescription, respectively.
- $P(A_1) = 0.5$ (Given: patient chooses one of two options with equal probability)
- $P(A_2) = 0.5$
The initial probability of a heart attack is $0.40$.
Step 2: Calculating Conditional Probabilities
We determine the reduced probability of a heart attack given the chosen course:
- If the patient chooses meditation ($A_1$), the risk is reduced by $30\%$: $P(E|A_1) = 0.40 \times (1 - 0.30) = 0.40 \times 0.70 = 0.28$ [By application of percentage reduction]
- If the patient chooses the drug ($A_2$), the risk is reduced by $25\%$: $P(E|A_2) = 0.40 \times (1 - 0.25) = 0.40 \times 0.75 = 0.30$ [By application of percentage reduction]
Step 3: Application of Bayes' Theorem
We are required to find $P(A_1|E)$, the probability that the patient followed meditation/yoga given they suffered a heart attack. [Per Bayes' Theorem: $P(A_i|E) = \frac{P(A_i)P(E|A_i)}{\sum P(A_j)P(E|A_j)}$]
Calculating the total probability $P(E)$:
$P(E) = P(A_1)P(E|A_1) + P(A_2)P(E|A_2)$ $P(E) = (0.5 \times 0.28) + (0.5 \times 0.30)$ $P(E) = 0.14 + 0.15 = 0.29$Step 4: Final Probability Computation
$P(A_1|E) = \frac{P(A_1)P(E|A_1)}{P(E)}$ $P(A_1|E) = \frac{0.5 \times 0.28}{0.29}$ $P(A_1|E) = \frac{0.14}{0.29} = \frac{14}{29}$Visual Representation of Probability Distribution
Final Solution: The probability that the patient followed a course of meditation and yoga is $\frac{14}{29} \approx 0.4827$.
More Questions from Class 12 Mathematics Probability Miscellaneous Exercise on Chapter 13
- Q1(i): $A$ and $B$ are two events such that $P(A) \neq 0$. Find $P(B|A)$, if (i) $A$ is a subset of $B$
- Q1(ii): $A$ and $B$ are two events such that $P(A) \neq 0$. Find $P(B|A)$, if (ii) $A \cap B = \phi$
- Q2(i): A couple has two children, (i) Find the probability that both children are males, if it is known that at least one of the children is male.
- Q2(ii): A couple has two children, (ii) Find the probability that both children are females, if it is known that the elder child is a female.
- Q3: Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
- Q4: Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
- Q5: If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?
- Q6: Suppose we have four boxes A,B,C and D containing coloured marbles as given below: One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?
- Q8: If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$).
- Q9(i): An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$, $P(B \text{ fails alone}) = 0.15$, $P(A \text{ and } B \text{ fail}) = 0.15$. Evaluate the following probabilities (i) $P(A \text{ fails|B has failed})$
- Q9(ii): An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known: $P(A \text{ fails}) = 0.2$, $P(B \text{ fails alone}) = 0.15$, $P(A \text{ and } B \text{ fail}) = 0.15$. Evaluate the following probabilities (ii) $P(A \text{ fails alone})$
- Q10: Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.
- Q11: If A and B are two events such that $P(A) \neq 0$ and $P(B | A) = 1$, then
- Q12: If $P(A|B) > P(A)$, then which of the following is correct :
- Q13: If A and B are any two events such that $P(A) + P(B) – P(A \text{ and } B) = P(A)$, then
CBSE Solutions for Class 12 Mathematics Probability
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