Advertisements
Advertisements
प्रश्न
A year is selected at random. What is the probability that it contains 53 Sundays
Advertisements
उत्तर
Probability of year being a leap year = `1/4`
Probability of year being non – leap year = `3/4`
A non – leap year has 365 days.
365 days = 52 weeks + 1 day.
52 weeks contain 52 Sundays.
In order to get 53 Sundays in a non – leap year the remaining I day must be a Sunday.
Remaining one day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday.
Probability of getting Sunday from the remaining one day = `1/7`
A leap year has 366 days.
366 days = 52 weeks + 2 odd days
52 weeks contain 52 Sundays.
In order to get 53 Sundays in a leap year the remaining 2 days must contain a Sunday.
Remaining Two days may be
S = (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday ), ( Friday, Saturday), (Saturday, Sunday)}
n(S) = 7
Let A be the event set of getting a Sunday then
A = {(Sunday, Monday), ( Saturday , Sunday)}
n(A) = 2
P(getting a Sunday from the remaining 2 days)
= `("n"("A"))/("n"("S"))`
= `2/7`
P(getting 53 Sundays in a year) = P(getting a leap year) × P(getting a Sunday from the remaining 2 days) + P(getting a non-leap year) × P(getting a Sunday from the remaining 1 day)
= `1/4 xx 2/7 + 3/4 xx 1/7`
= `2/28+ 3/28`
= `(2 + 3)/28`
= `5/28`
∴ Probability of getting 53 Sundays in a year = `5/28`
APPEARS IN
संबंधित प्रश्न
A die is thrown three times. Events A and B are defined as below:
A : 5 on the first and 6 on the second throw.
B: 3 or 4 on the third throw.
Find the probability of B, given that A has already occurred.
If P(A) = 0.8, P(B) = 0.5 and P(B|A) = 0.4, find P(A ∪ B)
Evaluate P(A ∪ B), if 2P(A) = P(B) = `5/13` and P(A | B) = `2/5`
Determine P(E|F).
A coin is tossed three times, where
E: at most two tails, F: at least one tail
A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5} Find P (E|G) and P (G|E)
A die is tossed thrice. Find the probability of getting an odd number at least once.
A and B are two events such that P (A) ≠ 0. Find P (B|A), if A is a subset of B.
A and B are two events such that P (A) ≠ 0. Find P (B|A), if A ∩ B = Φ.
In an examination, 30% of students have failed in subject I, 20% of the students have failed in subject II and 10% have failed in both subject I and subject II. A student is selected at random, what is the probability that the student has failed in at least one subject?
A bag contains 10 white balls and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that, first is white and second is black?
If A and B are two events such that P(A ∪ B) = 0.7, P(A ∩ B) = 0.2, and P(B) = 0.5, then show that A and B are independent
The probability that a car being filled with petrol will also need an oil change is 0.30; the probability that it needs a new oil filter is 0.40; and the probability that both the oil and filter need changing is 0.15. If a new oil filter is needed, what is the probability that the oil has to be changed?
Given P(A) = 0.4 and P(A ∪ B) = 0.7 Find P(B) if A and B are mutually exclusive
Two dice are thrown. Find the probability that the sum of numbers appearing is more than 11, is ______.
If P(A) = `4/5`, and P(A ∩ B) = `7/10`, then P(B|A) is equal to ______.
Two cards are drawn out randomly from a pack of 52 cards one after the other, without replacement. The probability of first card being a king and second card not being a king is:
A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is ______.
Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is ______.
If A and B are two events such that `P(A/B) = 2 xx P(B/A)` and P(A) + P(B) = `2/3`, then P(B) is equal to ______.
If P(B) = `3/5`, P(A | B) = `1/2` and P(A ∪ B) = `4/5`, then P(A ∪ B) + P(A' ∪ B) =
