Advertisements
Advertisements
प्रश्न
If P(A) = 0.5, P(B) = 0.8 and P(B/A) = 0.8, find P(A/B) and P(A ∪ B)
Advertisements
उत्तर
Given P(A) = 0.5
P(B) = 0.8
and P(B/A) = 0.8
P(A/B) = `("P"("A" ∩ "B"))/("P"("B"))` .........(1)
P(B/A) = `("P"("B" ∩ "A"))/("P"("A"))`
P(A ∩ B) = P(B/A) P(A)
Substituting in equation (1) we get
(1) ⇒ P(A/B) = `("P"("B"/"A") * "p"("A"))/("P"("B"))`
= `(0.8 xx 0.5)/0.8`
P(A/B) = 0.5
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) .........(2)
We have P(A/B) = `("P"("A" ∩ "B"))/("P"("B"))`
P(A ∩ B) = P(A/B) . P(B)
= 0.5 × 0.8
P(A ∩ B) = 0.40
(2) ⇒ P(A ∪ B) = 0.5 + 0.8 – 0.40
= 1.3 – 0.40
P(A ∪ B) = 0.90
APPEARS IN
संबंधित 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.
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find P(A ∪ B)
Determine P(E|F).
A coin is tossed three times, where
E: head on third toss, F: heads on first two tosses
Determine P(E|F).
A coin is tossed three times, where
E: at most two tails, F: at least one tail
Determine P(E|F).
Two coins are tossed once, where
E: tail appears on one coin, F: one coin shows head
A box has 20 pens of which 2 are defective. Calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.
In a college, 70% of students pass in Physics, 75% pass in Mathematics and 10% of students fail in both. One student is chosen at random. What is the probability that:
(i) He passes in Physics and Mathematics?
(ii) He passes in Mathematics given that he passes in Physics.
(iii) He passes in Physics given that he passes in Mathematics.
In an examination, 30% of students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subject I and subject II. A student is selected at random, what is the probability that the student has failed in at least one subject?
An urn contains 4 black, 5 white, and 6 red balls. Two balls are drawn one after the other without replacement, What is the probability that at least one ball is black?
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 black
Given P(A) = 0.4 and P(A ∪ B) = 0.7 Find P(B) if P(B/A) = 0.5
Choose the correct alternative:
If A and B are any two events, then the probability that exactly one of them occur is
If X denotes the number of ones in five consecutive throws of a dice, then P(X = 4) is ______
If P(A ∩ B) = `7/10` and P(B) = `17/20`, then P(A|B) equals ______.
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 3 red and 4 white balls and another bag contains 2 red and 3 white balls. If one ball is drawn from the first bag and 2 balls are drawn from the second bag, then find the probability that all three balls are of the same colour.
If for any two events A and B, P(A) = `4/5` and P(A ∩ B) = `7/10`, 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)
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?
Which conditional probabilities are equal to 1 when \[P(F)\neq 0\]?

