Question

In a leap year what is the probability of $53$ Sundays.

Answer 1

Hint: First consider $366$ days in a leap which can be written as $52$ weeks $+2$ extra days. Now find probable outcomes for two days and then find probability for which Sunday occurs.

Complete step-by-step answer:
In the question we are given a leap year and we have to find the probability of getting $53$ Sundays.
At first we will define what probability is and understand the basic terms related to the probability to be used in the question.
The probability of an event is a measure of the likelihood that the event would occur.
If an experiment’s outcomes are equally likely to occur, then the probability of an event E is the number of outcomes in E divided by number outcomes in the sample space. Here sample space consists of all the events that can occur possibly. It can be written as,
 $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Here $P\left( E \right)$ is probability of an event or an event which is asked $n\left( E \right)$ is number of favorable events and $n\left( S \right)$ is the number of all the events that can occur possibly.
Now let’s consider a leap year which contains $366$ days which means there are $52$ weeks $+2$ days extra
In this we can say that there are definitely $52$ Sundays which is true for other days too.
Now either of two days will be when the condition of $53$ Sundays is specified, so these two days can be
{Monday, Tuesday}, {Tuesday, Wednesday}, {Wednesday, Thursday}, {Thursday, Friday}, {Friday, Saturday}, {Saturday, Sunday}, {Sunday, Monday}
Total number of outcomes is $7$.
Now out of these outcomes only favorable are $2$ which is {Saturday, Sunday}, {Sunday, Monday} as they are having Sunday in them.
Hence the desired probability is $\dfrac{2}{7}$.

Note: If it would be a common or non-leap year there would only be one extra day so probability will be $\dfrac{1}{7}$. Students often make mistakes in these cases.