Advertisements
Advertisements
Question
If A and B are independent events, then A′ and B′ are also independent
Options
True
False
Advertisements
Solution
This statement is True.
Explanation:
Given A and B are independent.
∴ P(A ∩ B) = P(A)·P(B)
P(A' ∩ B') = P(A' ∪ B')
= 1 – P(A ∪ B)
= 1 – [P(A) + P(B) – P(A ∩ B)]
= 1 – P(A) – P(B) – P(A)·P(B)
= (1 – P(A))(1 – P(B))
= P(A')P(B')
Thus A and B are also independent.
APPEARS IN
RELATED QUESTIONS
A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A?
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).
Given that the events A and B are such that `P(A) = 1/2, PA∪B=3/5 and P (B) = p`. Find p if they are
- mutually exclusive
- independent.
Events A and B are such that `P(A) = 1/2, P(B) = 7/12 and P("not A or not B") = 1/4` . State whether A and B are independent?
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are `1/3`, `1/4`, and `1/5` respectively. If all of them try independently, what is the probability that, problem is solved?
A problem in statistics is given to three students A, B, and C. Their chances of solving the problem are `1/3`, `1/4`, and `1/5` respectively. If all of them try independently, what is the probability that, exactly two students solve the problem?
The odds against a husband who is 55 years old living till he is 75 is 8: 5 and it is 4: 3 against his wife who is now 48, living till she is 68. Find the probability that the couple will be alive 20 years hence.
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?
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(B)
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")`
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)
Solve the following:
Consider independent trails consisting of rolling a pair of fair dice, over and over What is the probability that a sum of 5 appears before sum of 7?
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.
If A and B′ are independent events then P(A′ ∪ B) = 1 – ______.
The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate `"P"(bar"A") + "P"(bar"B")`
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.
A and B are two events such that P(A) = `1/2`, P(B) = `1/3` and P(A ∩ B) = `1/4`. Find: `"P"("A"/"B")`
A and B are two events such that P(A) = `1/2`, P(B) = `1/3` and P(A ∩ B) = `1/4`. Find: `"P"("A'"/"B'")`
If A and B are two events such that P(A) = `1/2`, P(B) = `1/3` and P(A/B) = `1/4`, P(A' ∩ B') equals ______.
A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5. Then P(B′ ∩ A) 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 mutually exclusive events, then they will be independent also.
Two independent events are always mutually exclusive.
If A and B are two independent events then P(A and B) = P(A).P(B).
If A and B′ are independent events, then P(A' ∪ B) = 1 – P (A) P(B')
If A and B are two events such that P(A|B) = p, P(A) = p, P(B) = `1/3` and P(A ∪ B) = `5/9`, then p = ______.
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 black’
F : ‘the card drawn is a king’
Two events 'A' and 'B' are said to be independent if
Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Let A and B be independent events P(A) = 0.3 and P(B) = 0.4. Find P(A ∩ B)
The probability that A hits the target is `1/3` and the probability that B hits it, is `2/5`. If both try to hit the target independently, find the probability that the target is hit.
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 ______.
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 ______.
Two players A and B are alternately throwing a coin and a die together. A player who first throws head and 6 wins the game. If A starts the game, then the probability that B wins the game is ______.
Let E and F be two independent events. The probability that both E and F happen is `1/12` and the probability that neither E nor F happens is `1/2`, then a value of `(P(E))/(P(F))` is ______.
