# Solve the following: Find the probability that a year selected will have 53 Wednesdays - Mathematics and Statistics

Sum

Solve the following:

Find the probability that a year selected will have 53 Wednesdays

#### Solution

Let A ≡ the event that a year selected has 53 Wednesdays

L ≡ the event that leap year is selected

N ≡ the event that non-leap year is selected

The required event will happen if any one of L ∩ A and N ∩ A occurs.

These events are mutually exclusive

∴ the required probability

= P(L ∩ A) + P(N ∩ A)

= "P"("L")*"P"("A"/"L") + "P"("N")*"P"("A"/"N")  ...(1)

There is one leap year in 4 consecutive years

∴ P(L) = 1/4

"P"("A"/"L") = Probability that year has 53 Wednesdays given that it is a leap year

Leap year has 366 day

366 = 7 x 52 + 2

∴ there are 52 full weeks and 2 days.

These days can be Sunday, Monday; Monday, Tuesday; Tuesday, Wednesday; Wednesday, Thursday; Thursday, Friday; Friday, Saturday; Saturday, Sunday.

There are 7 possibilities and favourable cases are 2

∴ "P"("A"/"L") = 2/7

Since there are 3 non-leap years in 4 consecutive years,

P(N) = 3/4

"P"("A"/"N") = Probability that year has 53 Wednesdays given that it is a non-leap year

Non-leap year has 365 days

365 = 7 x 52 + 1

∴ there are 52 full weeks and 1 day.

This day could be any day of the week days i.e., any one of 7 days.

The number of favourable case is 1.

∴ "P"("A"/"N") = 1/7

∴ from (1), the required probability = 1/4*2/7 + 3/4*1/7

= 5/28.

Concept: Independent Events
Is there an error in this question or solution?

#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 9 Probability
Miscellaneous Exercise 9 | Q II. (13) | Page 214