Advertisements
Advertisements
प्रश्न
The probability of simultaneous occurrence of at least one of two events A and B is p. If the probability that exactly one of A, B occurs is q, then prove that P(A′) + P(B′) = 2 – 2p + q.
Advertisements
उत्तर
Since P(exactly one of A, B occurs) = q .....(Given)
We get P(A ∪ B) – P(A ∩ B) = q
⇒ p – P(A ∩ B) = q
⇒ P(A ∩ B) = p – q
⇒ 1 – P(A′ ∪ B′) = p – q
⇒ P(A′ ∪ B′) = 1 – p + q
⇒ P(A′) + P(B′) – P(A′ ∩ B′) = 1 – p + q
⇒ P(A′) + P(B′) = (1 – p + q) + P(A′ ∩ B′)
= (1 – p + q) + (1 – P(A ∪ B))
= (1 – p + q) + (1 – p)
= 2 – 2p + q.
APPEARS IN
संबंधित प्रश्न
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 `1/2`).
Prove that if E and F are independent events, then the events E and F' are also independent.
The probability that a 50-year old man will be alive till age 60 is 0.83 and the probability that a 45-year old woman will be alive till age 55 is 0.97. What is the probability that a man whose age is 50 and his wife whose age is 45 will both be alive after 10 years?
One-shot is fired from each of the three guns. Let A, B, and C denote the events that the target is hit by the first, second and third guns respectively. assuming that A, B, and C are independent events and that P(A) = 0.5, P(B) = 0.6, and P(C) = 0.8, then find the probability that at least one hit is registered.
The odds against a certain event are 5: 2 and odds in favour of another independent event are 6: 5. Find the chance that at least one of the events will happen.
The odds against student X solving a business statistics problem are 8: 6 and odds in favour of student Y solving the same problem are 14: 16 What is the chance that the problem will be solved, if they try independently?
Two hundred patients who had either Eye surgery or Throat surgery were asked whether they were satisfied or unsatisfied regarding the result of their surgery
The follwoing table summarizes their response:
| Surgery | Satisfied | Unsatisfied | Total |
| Throat | 70 | 25 | 95 |
| Eye | 90 | 15 | 105 |
| Total | 160 | 40 | 200 |
If one person from the 200 patients is selected at random, determine the probability that the person was satisfied given that the person had Throat surgery.
The probability that a man who is 45 years old will be alive till he becomes 70 is `5/12`. The probability that his wife who is 40 years old will be alive till she becomes 65 is `3/8`. What is the probability that, 25 years hence,
- the couple will be alive
- exactly one of them will be alive
- none of them will be alive
- at least one of them will be alive
A bag contains 3 yellow and 5 brown balls. Another bag contains 4 yellow and 6 brown balls. If one ball is drawn from each bag, what is the probability that, the balls are of different color?
A bag contains 3 red and 5 white balls. Two balls are drawn at random one after the other without replacement. Find the probability that both the balls are white.
Solution: Let,
A : First ball drawn is white
B : second ball drawn in white.
P(A) = `square/square`
After drawing the first ball, without replacing it into the bag a second ball is drawn from the remaining `square` balls.
∴ P(B/A) = `square/square`
∴ P(Both balls are white) = P(A ∩ B)
`= "P"(square) * "P"(square)`
`= square * square`
= `square`
Solve the following:
If P(A ∩ B) = `1/2`, P(B ∩ C) = `1/3`, P(C ∩ A) = `1/6` then find P(A), P(B) and P(C), If A,B,C are independent events.
Solve the following:
Let A and B be independent events with P(A) = `1/4`, and P(A ∪ B) = 2P(B) – P(A). Find `"P"("A"/"B")`
Two dice are thrown together. Let A be the event ‘getting 6 on the first die’ and B be the event ‘getting 2 on the second die’. Are the events A and B independent?
If A and B are independent events such that P(A) = p, P(B) = 2p and P(Exactly one of A, B) = `5/9`, then p = ______.
Two dice are thrown together and the total score is noted. The events E, F and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are independent.
Let E1 and E2 be two independent events such that P(E1) = P1 and P(E2) = P2. Describe in words of the events whose probabilities are: (1 – P1) P2
If A and B are two events and A ≠ Φ, B ≠ Φ, then ______.
If A and B are such events that P(A) > 0 and P(B) ≠ 1, then P(A′|B′) equals ______.
If two events are independent, then ______.
If the events A and B are independent, then P(A ∩ B) is equal to ______.
Let P(A) > 0 and P(B) > 0. Then A and B can be both mutually exclusive and independent.
Two independent events are always mutually exclusive.
If A, B and C are three independent events such that P(A) = P(B) = P(C) = p, then P(At least two of A, B, C occur) = 3p2 – 2p3
Let A and B be two events. If P(A | B) = P(A), then A is ______ of B.
One card is drawn at random from a well-shuffled deck of 52 cards. In which of the following case is the events E and F independent?
E : ‘the card drawn is a spade’
F : ‘the card drawn is an ace’
The probability of obtaining an even prime number on each die when a pair of dice is rolled is
Two events 'A' and 'B' are said to be independent if
If P(A) = `3/5` and P(B) = `1/5`, find P(A ∩ B), If A and B are independent events.
Let A and B be independent events P(A) = 0.3 and P(B) = 0.4. Find P(A ∩ B)
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6 and P(A' ∩ B') is ______.
A problem in Mathematics is given to three students whose chances of solving it are `1/2, 1/3, 1/4` respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is ______.
The probability of the event A occurring is `1/3` and of the event B occurring is `1/2`. If A and B are independent events, then find the probability of neither A nor B occurring.
Given two events A and B such that (A/B) = 0.25 and P(A ∩ B) = 0.12. The value P(A ∩ B') is ______.
Two events E and F are independent when which conditional probability condition holds, provided \[P(E) \ne 0\]?
For \[E=\{3,6\}\] and \[F=\{2,4,6\}\], what is \[E\cap F\]?
For \[F=\{2,4,6\}\] when a die is thrown once, what is \[P(F)\]?
For \[E\cap F=\{6\}\] when a die is thrown once, what is \[P(E\cap F)\]?
Given \[P(E\cap F)=\frac{1}{6}\] and \[P(E)P(F)=\frac{1}{6}\], what is the relation between E and F?
