Advertisements
Advertisements
प्रश्न
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 and yoga. Interpret the result and state which of the above stated methods is more beneficial for the patient.
Advertisements
उत्तर
Let A, E1, and E2 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.
`therefore P(A)=0.40`
`P(E_1)=P(E_2)=1/2`
`P(A|E_1)=0.40xx0.70=0.28P(A|E_2)=0.40xx0.75=0.30`
Probability that the patient suffering a heart attack followed a course of meditation and yoga is given by `P (E1|A).`
`P(E_1|A)=(P(E_1)P(A|E_1))/(P(E_1)P(A|E_1)+P(E_2)P(A|E_2))`
`=(1/2xx0.28)/(1/2xx0.28+1/2xx0.30)`
`=14/29`
Let us calculate `P(E_2|A)`
`P(E_2|A)=(P(E_2)P(A|E_2))/(P(E_1)P(A|E_1)+P(E_2)P(A|E_2))`
`=(1/2xx0.30)/(1/2xx0.28+1/2xx0.30)`
`=15/29`
Since `P(E_1|A)< P(E_2|A) ` the course of yoga and meditation is more beneficial for a person having chances of heart attack.
APPEARS IN
संबंधित प्रश्न
Given that the two numbers appearing on throwing the two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.
If P(A) = `1/2`, P(B) = 0, then P(A|B) is ______.
A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
If events A and B are independent, such that `P(A)= 3/5`, `P(B)=2/3` 'find P(A ∪ B).
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that, one is white and other is black?
Two cards are drawn one after the other from a pack of 52 cards without replacement. What is the probability that both the cards drawn are face cards?
Select the correct option from the given alternatives :
Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. The probability that it was drawn from Bag II
If A and B are two events such that P(A ∪ B) = 0.7, P(A ∩ B) = 0.2, and P(B) = 0.5, then show that A and B are independent
One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that both are white
One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that one white and one black
Two dice are thrown. Find the probability that the sum of numbers appearing is more than 11, is ______.
Three machines E1, E2, E3 in a certain factory produced 50%, 25% and 25%, respectively, of the total daily output of electric tubes. It is known that 4% of the tubes produced one each of machines E1 and E2 are defective, and that 5% of those produced on E3 are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.
If P(A) = `3/10`, P(B) = `2/5` and P(A ∪ B) = `3/5`, then P(B|A) + P(A|B) equals ______.
Two cards are drawn out randomly from a pack of 52 cards one after the other, without replacement. The probability of first card being a king and second card not being a king is:
If two balls are drawn from a bag containing 3 white, 4 black and 5 red balls. Then, the probability that the drawn balls are of different colours is:
A bag contains 6 red and 5 blue balls and another bag contains 5 red and 8 blue balls. A ball is drawn from the first bag and without noticing its colour is placed in the second bag. If a ball is drawn from the second bag, then find the probability that the drawn ball is red in colour.
If the sum of numbers obtained on throwing a pair of dice is 9, then the probability that number obtained on one of the dice is 4, is ______.
If for two events A and B, P(A – B) = `1/5` and P(A) = `3/5`, then `P(B/A)` is equal to ______.
Read the following passage:
|
Recent studies suggest the roughly 12% of the world population is left-handed.
Assuming that P(A) = P(B) = P(C) = P(D) = `1/4` and L denotes the event that child is left-handed. |
Based on the above information, answer the following questions:
- Find `P(L/C)` (1)
- Find `P(overlineL/A)` (1)
- (a) Find `P(A/L)` (2)
OR
(b) Find the probability that a randomly selected child is left-handed given that exactly one of the parents is left-handed. (2)
A Problem in Mathematics is given to the three students A, B and C. Their chances of solving the problem are `1/2, 1/3` and `1/4` respectively. Find the probability that at least two of them will solve the problem.
Three friends go to a restaurant to have pizza. They decide who will pay for the pizza by tossing a coin. It is decided that each one of them will toss a coin and if one person gets a different result (heads or tails) than the other two, that person would pay. If all three get the same result (all heads or all tails), they will toss again until they get a different result.
- What is the probability that all three friends will get the same result (all heads or all tails) in one round of tossing?
- What is the probability that they will get a different result in one round of tossing?
- What is the probability that they will need exactly four rounds of tossing to determine who would pay?
Students of under graduation submitted a case study on “Understanding the Probability of Left-Handedness in Children Based on Parental Handedness”. Following Recent studies suggest that roughly 12% of the world population is left-handed. Depending on the parents’ handedness, the chances of having a left-handed child are as follows:
Scenario A: Both parents are left-handed, with a 24% chance of the child being left-handed.
Scenario B: The fathers is right-handed and the mothers left-handed, with a 22% chance of child being left-handed.
Scenario C: The fathers left-handed and the mother is right-handed, with a 17% chance of child being left-handed.
Scenario D: Both parents are right-handed, with a 9% chance of having a left-handed child.
Assuming that scenarios A, B, C and D are equally likely and L denotes the event that the child is left-handed, answer the following questions.
- What is the overall probability that a randomly selected child is left-handed?
- Given that exactly one parent is left-handed, what is the probability that a randomly selected child is left-handed?
- If a child is left-handed, what is the probability that both parents are left-handed?
If A and B are events and \[P(B) \neq 0\], which expression gives \[P(A \mid B)\]?
Which statement correctly describes the numerator in \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}\]?
For independent events, which equality holds provided \[P(B)\neq 0\]?

