Question

Find the probability that there are $53$ Mondays in a
A. Leap year
B. Non-leap year

Answer 1

We will first write down the no. of days in a year and then find out the no. of weeks and multiply by 7 to get the no. of days then we will it subtract it from $366$ for leap year and from $365$ for a non-leap year and then we have to see the possibilities for the leftover days and see all the outcomes and separate the favourable outcomes (that is one with the Mondays) and finally apply the formula for $\text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$ .

Complete step-by-step answer :
We know that the probability for an event to happen is:
$\text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
Let’s consider the first part that is the probability of $53$ Mondays in a Leap year :
Now we know that a year has $365$ days whereas a leap year has $366$ days.
Now a normal year has $52$ weeks that means there will be $52$ Mondays for sure.
Now $1$ week has $7$ days therefore $52$ day has = $52\times 7=364$ days.
Now, we know that a leap year has $366$ days and $366-364=2$ , means in a leap year there will be $52$ Mondays and $2$ days will be left.
Now, these two days can be as follows:
Sunday, Monday
Monday, Tuesday
Tuesday, Wednesday
Wednesday, Thursday
Thursday, Friday
Friday, Saturday
Saturday, Sunday
Now, out of these total $7$ outcomes we need only those outcomes which has Mondays in it , therefore the favourable outcomes are $2$
Hence, the probability of getting $53$ Mondays in a leap year =
$\text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\dfrac{2}{7}$
Now let’s look at the second part that is the probability of $53$ Mondays in a Non-leap year:
Now we know that a year has $365$ days, now a normal year has $52$ weeks that means there will be $52$ Mondays for sure.
Now $1$ week has $7$ days therefore $52$ day has = $52\times 7=364$ days.
Now, we know that a non-leap year has $365$ days and $365-364=1$ , means in a non-leap year there will be $52$ Mondays and $1$ day will be left.
Now, these two days can be as follows:
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Now, out of these total $7$ outcomes we need only those outcomes which has Mondays in it , therefore the favourable outcomes are $1$
Hence, the probability of getting $53$ Mondays in a non-leap year =
$\text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\dfrac{1}{7}$
Hence, the probability of getting $53$ Mondays in
A. leap year = $\dfrac{2}{7}$
B. non-leap year = $\dfrac{1}{7}$

Note : Always describe in detail while solving the probability questions to help the examiner to understand your solution. Students might get confused when we say that a year has 52 weeks as we can see that dividing $\dfrac{365}{7}=52.14$ but we normally ignore the decimal point.