What is the probability of having 53 thursday in a leap year

What is the probability of having 53 Thursdays in a leap year?

Asked by Topperlearning User | 04 Apr, 2014, 10:52: AM

In a leap year, there are 366 days.

52 weeks = 364 days

1 leap year = 52 weeks and 2 days

These extra two days can be sun-mon, mon-tue, tue-wed, wed-thu, thu-fir, fri-sat, sat-sun.

Total number of outcomes = 7

Number of favorable outcomes = 2

P(53 Thursdays) =

Answered by | 04 Apr, 2014, 12:52: PM

getcalc.com's Probability calculator to find what is the probability of 53 Thursdays in a leap year. The ratio of expected event to all the possible events of a sample space for 2 odd days to be either {Wednesday & Thursday} or {Thursday & Friday} is the probability of getting 53 Thursdays for a leap year.
P(A) = 2/7 = 0.28

Users may refer the below detailed information to learn how to find the probability of 53 Thursdays in a leap year. The total number of weeks in a non-leap year {366 days = 52 (2/7)} is 52 weeks and two odd days. Since, finding the probability for the two odd days to be either {Wednesday & Thursday} or {Thursday & Friday} is enough to find the probability of getting 53 Thursdays in an ordinary year of a Gregorian calendar.

Workout
step 1 Possible events for 2 odd days The odd day may be either Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday. Therefore, the total number of possible outcome or elements of sample space is 7.

step 2 Probability of 2 Odd day to be Thursday :

P(A) = S/A={{Wednesday & Thursday}, {Thursday & Friday}}/{Saturday & Sunday, . . . , Friday & Saturday}

0.28 or 2/7 is probability for 53 Thursdays in a leap year.

Class-X Maths Probability

What is the probability of getting 53 thursdays in a leap year?