Question

What are the chances a leap year selected at random will contain 53 sundays?
A. The required probability is \[\dfrac{2}{7}\]
B. The required probability is \[\dfrac{3}{7}\]
C. The required probability is \[\dfrac{1}{7}\]
D. The required probability is \[\dfrac{4}{7}\]

Answer 1

Hint: A leap year has 366 days and a normal year has 365 days, A Year contains 52 complete weeks and a week has 7 days so if we multiply 7 with 52 we will get 364 which means we have one spare day in normal years and 2 spare days in a leap year.

Complete step-by-step answer:
According to the hint it is clear that there are a total of 52 sundays in 52 weeks.. Now the \[{53^{rd}}\] sunday will have to only if sundays are present in those two spare days. So for this question first we will write all the possibilities of those 2 days and then we will choose how many times sunday comes in those possibilities and then after dividing the total number of time sunday comes by total number of favourable outcomes we get the required probability
So the total outcomes are as follows {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday},{Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}. Now if we observe closely it can be noted that sunday only occurs 2 times in all the given sets of possible outcomes and there are a total of 7 favourable outcomes. So from here only we can say that the required probability is \[\dfrac{2}{7}\]
Hence option A is the correct option here.


Note: It must be noted that the probability of anything is always the number of required outcomes divided by the total number of favourable outcomes, this formula holds until and unless conditional probability comes into play, where 2 or more events happen simultaneously.