Question

What is the chance that non-leap year

Answer 1

No of days = 365

= `365/7` weeks

= 52 weeks + 1 day

In 52 weeks we have 52 Sundays.

So we have to find the probability of getting the remaining one day as Sunday.

The remaining 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday

(i.e.,) n(S) = 7

In the n (Sunday) = A {Saturday to Sunday or Sunday to Monday}

(i.e.,) n(A) = 1

So, P(A) = 1.

∴ Probability of getting 53 Sundays `("n"("A"))/("n"("S")) = 1/7`