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Tamil Nadu Board of Secondary EducationHSC Science Class 11

A year is selected at random. What is the probability that it contains 53 Sundays - Mathematics

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Question

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

Sum
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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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Chapter 12: Introduction to probability theory - Exercise 12.3 [Page 259]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 12 Introduction to probability theory
Exercise 12.3 | Q 11. (i) | Page 259

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