Advertisements
Advertisements
प्रश्न
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?
उत्तर
A speaks truth `P(A) = 60/100`, `P(A') = 40/100`
B speaks truth `P(B) = 40/100`, `P(B') = 60/100`
they contradict each other = `P(A).(B') + P(A') . P(B)`
`= 60/10 xx 60/100 + 40/100 xx 40/100`
`= (3600+1600)/10000`
`= 5200/10000`
`= 52/100`
% of cases they likey to contradict each other = `52/100 xx 100 = 52%`
APPEARS IN
संबंधित प्रश्न
A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all balls in the bag are white?
If `P(A) = 3/5 and P(B) = 1/5` , find P (A ∩ B) if A and B are independent events.
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.
If P(A) = 0·4, P(B) = p, P(A ⋃ B) = 0·6 and A and B are given to be independent events, find the value of 'p'.
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?
An urn contains four tickets marked with numbers 112, 121, 122, 222 and one ticket is drawn at random. Let Ai (i = 1, 2, 3) be the event that ith digit of the number of the ticket drawn is 1. Discuss the independence of the events A1, A2, and A3.
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?
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.
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, both the balls are of the same color?
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?
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:
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"("B'"/"A'")`
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.
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"("A'"/"B'")`
Let A and B be two events such that P(A) = `3/8`, P(B) = `5/8` and P(A ∪ B) = `3/4`. Then P(A|B).P(A′|B) is equal to ______.
If A and B are mutually exclusive events, then they will be independent also.
If A and B are independent, then P(exactly one of A, B occurs) = P(A)P(B') + P(B)P(A')
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.
