Question

What is the probability that there are 53 sundays and 53 saturdays in a leap year?

Answer 1

Answer:

Step-by-step explanation:

Leap year has 366 days in which 52weeks + 2 extra days

366= 52*7+2

Those 2 days can be any 2 successive days {sunday-monday, mon-tues, tues-wednes, wednes-thurs, thurs-fri, fri-sat, sat-sun}

In the question given we need the probability of 53 sundays and 53 saturdays so there is only one favourable possibility out of seven possibilities.

Ans : 1/7

Answer 2

The probability that there are 53 sundays and 53 saturdays in a leap year = 1/7

Given :

A leap year

To find :

The probability that there are 53 sundays and 53 saturdays in a leap year

Solution :

Step 1 of 3 :

Find total number of possible outcomes

Leap year = 366 days

366 days = 52 weeks 2 days

Now 52 weeks contains 52 Sundays

2 days is one of the below

( Sunday, Monday), ( Monday, Tuesday), (Tuesday, Wednesday), ( Wednesday, Thursday), ( Thursday, Friday), ( Friday, Saturday ), (Saturday, Sunday)

So the total number of possible outcomes = 7

Step 2 of 3 :

Find total number of possible outcomes for the event

Let A be the event that there are 53 sundays and 53 saturdays in a leap year

So the total event points for the event A is (Saturday, Sunday)

So the total number of possible outcomes for the event A is 1

Step 3 of 3 :

Find the probability

Hence the required probability

= The probability that there are 53 sundays and 53 saturdays in a leap year

= P(A)

$\displaystyle \sf{ = \frac{Number \: of \: favourable \: cases \: to \: the \: event \: A }{Total \: number \: of \: possible \: outcomes }}$

$\displaystyle \sf{ = \frac{1}{7} }$

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