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Question
A year is selected at random. What is the probability that it contains 53 Sundays
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Solution
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`
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