Advertisements
Advertisements
Question
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
Options
`52/96`
`71/96`
`69/96`
`13/96`
Advertisements
Solution
`71/96`
Explanation;
[Hint : Required probability = `1 - 5/8 xx 5/12`
= `1 - 25/96 = 71/96`]
APPEARS IN
RELATED QUESTIONS
A card from a pack of 52 playing cards is lost. From the remaining cards of the pack three cards are drawn at random (without replacement) and are found to be all spades. Find the probability of the lost card being a spade.
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?
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?
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)
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 not solved
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?
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 person was unsatisfied given that the person had eye surgery
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 following 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 the person had Throat surgery given that the person was unsatisfied.
Bag A contains 3 red and 2 white balls and bag B contains 2 red and 5 white balls. A bag is selected at random, a ball is drawn and put into the other bag, and then a ball is drawn from that bag. Find the probability that both the balls drawn are of same 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:
If P(A) = `"P"("A"/"B") = 1/5, "P"("B"/"A") = 1/3` the find `"P"("A'"/"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(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"("B'"/"A")`
10% of the bulbs produced in a factory are of red colour and 2% are red and defective. If one bulb is picked up at random, determine the probability of its being defective if it is red.
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 = ______.
For a loaded die, the probabilities of outcomes are given as under:
P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1 and P(4) = 0.3. The die is thrown two times. Let A and B be the events, ‘same number each time’, and ‘a total score is 10 or more’, respectively. Determine whether or not A and B are independent.
Refer to Question 1 above. If the die were fair, determine whether or not the events A and B 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"("B"/"A")`
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 ______.
In Question 64 above, P(B|A′) is equal to ______.
If A and B are such events that P(A) > 0 and P(B) ≠ 1, 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 ______.
If the events A and B are independent, then P(A ∩ B) is equal to ______.
If A and B are two independent events then P(A and B) = P(A).P(B).
Let A and B be two events. If P(A | B) = P(A), then A is ______ of B.
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)
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.
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 ______.
A die is thrown once. What is the sample space \[S\]?
Given \[P(E\cap F)=\frac{1}{6}\] and \[P(E)P(F)=\frac{1}{6}\], what is the relation between E and F?
