Advertisements
Advertisements
प्रश्न
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 at least one of them will be alive 20 years hence.
उत्तर
Let A be the event that husband would be alive after 20 years.
Odds against A are 8: 5
∴ The probability of occurrence of event A is given by
P(A) = `5/(8 + 5) = 5/13`
∴ P(A') = 1 – P(A) = `1 - 5/13 = 8/13`
Let B be the event that wife would be alive after 20 years.
Odds against B are 4: 3
∴ The probability of occurrence of event B is given by
P(B) = `3/(4 + 3) = 3/7`
∴ P(B') = 1 – P(B) = `1 - 3/7 = 4/7`
Since A and B are independent events
∴ A' and B' are also independent events
Let Y be the event that at least one will be alive after 20 years.
∴ P(Y) = P(at least one would be alive)
= 1 – P(both would not be alive)
= 1 – P(A' ∩ B')
= 1 – P(A') . P(B')
= `1 - 8/13 xx 4/7`
= `1 - 32/91`
= `59/91`
APPEARS IN
संबंधित प्रश्न
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.
Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find
- P (A ∩ B)
- P (A ∪ B)
- P (A | B)
- P (B | A)
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?
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.
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:
Let A and B be independent events with P(A) = `1/4`, and P(A ∪ B) = 2P(B) – P(A). Find `"P"("B'"/"A")`
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?
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).
Refer to Question 1 above. If the die were fair, determine whether or not the events A and B are independent.
Three events A, B and C have probabilities `2/5, 1/3` and `1/2`, , respectively. Given that P(A ∩ C) = `1/5` and P(B ∩ C) = `1/4`, find the values of P(C|B) and P(A' ∩ C').
If A and B are two events such that P(B) = `3/5`, P(A|B) = `1/2` and P(A ∪ B) = `4/5`, then P(A) equals ______.
If A and B are independent events, then A′ and B′ are also independent
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, B are two events such that `1/8 ≤ P(A ∩ B) ≤ 3/8` then
If P(A) = `3/5` and P(B) = `1/5`, find P(A ∩ B), If A and B are independent events.
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 ______.
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.
