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Question
What is the chance that non-leap year
Solution
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`
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