Advertisements
Advertisements
प्रश्न
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’
उत्तर
P(E) = `13/52 = 1/4`
P(F) = `4/52 = 1/13`
P(E ∩ F) = `1/52`
P(E) . P(F) = `1/4 xx 1/13 = 1/52` = P(E ∩ F)
Therefore E and F are independent events.
APPEARS IN
संबंधित प्रश्न
If A and B are two independent events such that `P(barA∩ B) =2/15 and P(A ∩ barB) = 1/6`, then find P(A) and P(B).
A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event "number obtained is even" and B be the event "number obtained is red". Find if A and B are independent events.
Let E and F be events with `P(E) = 3/5, P(F) = 3/10 and P(E ∩ F) = 1/5`. Are E and F independent?
If A and B are two events such that `P(A) = 1/4, P(B) = 1/2 and P(A ∩ B) = 1/8`, find P (not A and not B).
Given two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find
- P (A and B)
- P(A and not B)
- P(A or B)
- P(neither A nor B)
Probability of solving specific problem independently by A and B are `1/2` and `1/3` respectively. If both try to solve the problem independently, find the probability that
- the problem is solved
- exactly one of them solves the problem.
Two events, A and B, will be independent if ______.
A speaks the truth in 60% of the cases, while B is 40% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact?
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?
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 probability that neither solves the problem?
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`
A family has two children. Find the probability that both the children are girls, given that atleast one of them is a girl.
Select the correct option from the given alternatives :
The odds against an event are 5:3 and the odds in favour of another independent event are 7:5. The probability that at least one of the two events will occur is
Solve the following:
Find the probability that a year selected will have 53 Wednesdays
Solve the following:
A and B throw a die alternatively till one of them gets a 3 and wins the game. Find the respective probabilities of winning. (Assuming A begins the game)
If A and B are independent events such that 0 < P(A) < 1 and 0 < P(B) < 1, then which of the following is not correct?
If A and B′ are independent events then P(A′ ∪ B) = 1 – ______.
Let A and B be two independent events. Then P(A ∩ B) = P(A) + P(B)
Three events A, B and C are said to be independent if P(A ∩ B ∩ C) = P(A) P(B) P(C).
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 – (1 – P1)(1 – P2)
If A and B are two events and A ≠ Φ, B ≠ Φ, then ______.
If A and B are two independent events with P(A) = `3/5` and P(B) = `4/9`, then P(A′ ∩ B′) equals ______.
If A and B are two independent events with P(A) = `3/5` and P(B) = `4/9`, then P(A′ ∩ B′) equals ______.
Let P(A) > 0 and P(B) > 0. Then A and B can be both mutually exclusive and independent.
If A and B are independent events, then A′ and B′ are also independent
If A and B are two events such that P(A) > 0 and P(A) + P(B) >1, then P(B|A) ≥ `1 - ("P"("B'"))/("P"("A"))`
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
Two events 'A' and 'B' are said to be independent if
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 ______.
Events A and Bare such that P(A) = `1/2`, P(B) = `7/12` and `P(barA ∪ barB) = 1/4`. Find whether the events A and B are independent or not.
Let Bi(i = 1, 2, 3) be three independent events in a sample space. The probability that only B1 occur is α, only B2 occurs is β and only B3 occurs is γ. Let p be the probability that none of the events Bi occurs and these 4 probabilities satisfy the equations (α – 2β)p = αβ and (β – 3γ) = 2βy (All the probabilities are assumed to lie in the interval (0, 1)). Then `("P"("B"_1))/("P"("B"_3))` is equal to ______.
Let EC denote the complement of an event E. Let E1, E2 and E3 be any pairwise independent events with P(E1) > 0 and P(E1 ∩ E2 ∩ E3) = 0. Then `"P"(("E"_2^"C" ∩ "E"_3^"C")/"E"_1)` is equal to ______.
Five fair coins are tossed simultaneously. The probability of the events that at least one head comes up is ______.
