Sum

What is the probability that an ordinary year has 53 Sundays?

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#### 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`

Concept: Probability - A Theoretical Approach

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