Question

find the probability of getting 53 Wednesdays in a leap year

Answer 1

A year= 365 days.
A leap year= 366 days.
A year has fifty two weeks. Hence there will be fifty two Wednesdays for sure.
fifty two weeks = 52×7= 364 days.
366-364=2.
in a leap year there will be fifty two Wednesdays and two days will be left.
These two days can be:
Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday.
Of these total 7 outcomes, the favourable out comes are 2.
Hence the probability of getting 53 Wednesdays in a leap year=2/7.

Answer 2

Answer:

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Step-by-step explanation:

Let P(E) be the event of probability of getting 53 wednesdays

A leap year consist of 366 days

365 = 52 × 7= 364 days and 2  days left from the complete weeks

This two days can be (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed,Thu), (Thu, Fri) (Fri,Sat) and (Sat, Sun)

Out of these 7 pairs, 2 of them have wednesday

So, number of favourable outcome to the event P(E) = 2

Total outcomes = 7

The probability of getting 53 mondays= 2/ 7

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