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Question
What is the probability that an ordinary year has 53 Sundays?
Solution 1
Ordinary year has 365 days
365 days = 52 weeks + 1 day
That 1 day may be Sun, Mon, Tue, Wed, Thu, Fri, Sat
Total no. of possible outcomes = 7
Let E ⟶ event of getting 53 Sundays
No. of favourable outcomes = 1 {Sun}
P(E) =`"No.of favorable outcomes"/"Total no.of possible outcomes"`=1/7
Solution 2
There are 365 days i.e., 52 weeks and 1 day in an ordinary year.
For the remaining day, the sample space is
S = {Mon., Tue., Wed., Thur., Fri., Sat., Sun.}
∴ n(S) = 7
Let A be the event that the remaining day is a Sunday.
∴ A = {Sunday}
∴ n(A) = 1
∴ P(A) = `("n"("A"))/("n"("S"))`
= `1/7`
∴ The probability that an ordinary year has 53 Sundays is `1/7`
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