Advertisements
Advertisements
प्रश्न
Solve the following:
Find the probability that a year selected will have 53 Wednesdays
Advertisements
उत्तर
Let A ≡ the event that a year selected has 53 Wednesdays
L ≡ the event that leap year is selected
N ≡ the event that non-leap year is selected
The required event will happen if any one of L ∩ A and N ∩ A occurs.
These events are mutually exclusive
∴ the required probability
= P(L ∩ A) + P(N ∩ A)
= `"P"("L")*"P"("A"/"L") + "P"("N")*"P"("A"/"N")` ...(1)
There is one leap year in 4 consecutive years
∴ P(L) = `1/4`
`"P"("A"/"L")` = Probability that year has 53 Wednesdays given that it is a leap year
Leap year has 366 day
366 = 7 x 52 + 2
∴ there are 52 full weeks and 2 days.
These days can be Sunday, Monday; Monday, Tuesday; Tuesday, Wednesday; Wednesday, Thursday; Thursday, Friday; Friday, Saturday; Saturday, Sunday.
There are 7 possibilities and favourable cases are 2
∴ `"P"("A"/"L") = 2/7`
Since there are 3 non-leap years in 4 consecutive years,
P(N) = `3/4`
`"P"("A"/"N")` = Probability that year has 53 Wednesdays given that it is a non-leap year
Non-leap year has 365 days
365 = 7 x 52 + 1
∴ there are 52 full weeks and 1 day.
This day could be any day of the week days i.e., any one of 7 days.
The number of favourable case is 1.
∴ `"P"("A"/"N") = 1/7`
∴ from (1), the required probability = `1/4*2/7 + 3/4*1/7`
= `5/28`.
APPEARS IN
संबंधित प्रश्न
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.
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.
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 ______.
Prove that if E and F are independent events, then the events E and F' are also independent.
In a race, the probabilities of A and B winning the race are `1/3` and `1/6` respectively. Find the probability of neither of them winning the race.
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?
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?
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.
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?
Solve the following:
For three events A, B and C, we know that A and C are independent, B and C are independent, A and B are disjoint, P(A ∪ C) = `2/3`, P(B ∪ C) = `3/4`, P(A ∪ B ∪ C) = `11/12`. Find P(A), P(B) and P(C)
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")`
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: P1P2
If two events are independent, then ______.
Two events E and F are independent. If P(E) = 0.3, P(E ∪ F) = 0.5, then P(E|F) – P(F|E) equals ______.
If A and B are two independent events then P(A and B) = P(A).P(B).
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"))`
Let A and B be two events. If P(A | B) = P(A), then A is ______ of B.
Let E1 and E2 be two independent events. Let P(E) denotes the probability of the occurrence of the event E. Further, let E'1 and E'2 denote the complements of E1 and E2, respectively. If P(E'1 ∩ E2) = `2/15` and P(E1 ∩ E'2) = `1/6`, then P(E1) is
If A, B are two events such that `1/8 ≤ P(A ∩ B) ≤ 3/8` then
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.
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)
Given two independent events, if the probability that exactly one of them occurs is `26/49` and the probability that none of them occurs is `15/49`, then the probability of more probable of the two events is ______.
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.
Given two events A and B such that (A/B) = 0.25 and P(A ∩ B) = 0.12. The value P(A ∩ B') 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?
