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Find the probability of getting 53 Fridays in a leap year.

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Hint- Here, we will be using the general formula for probability of occurrence of an event. Here, we will be finding out how exactly we will be able to obtain 53 Fridays in a leap year which consists of 52 weeks and 2 days.

As we know that in a leap year, there are 366 days and 7 days is equivalent to 1 week.
Clearly $366 = 364 + 2 = \left( {52 \times 7} \right) + 2 = $52 weeks and 2 days.
So, a leap year consists of 52 weeks and 2 days. In these 52 weeks, there will be 52 Fridays. Hence, the occurrence of 53 Fridays will completely depend on the 2 days left.
According to the general formula for probability in given by
Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of possible outcomes}}}}{\text{ }} \to {\text{(1)}}$
Here, the favourable event is the occurrence of Friday in any of the two days left.
All the possible cases of the remaining two days are Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday, Sunday and Monday.
Total number of possible outcomes=7
Here, favourable outcomes will count only those cases out of all the possible outcomes in which there is Friday occurring i.e., Thursday and Friday, Friday and Saturday only.
Number of favourable outcomes=2
Using the formula given in equation (1), we get
Probability of getting 53 Fridays in a leap year$ = \dfrac{{\text{2}}}{{\text{7}}}$.

Note- In these types of problems, we find out how many exact weeks are there (in this case it is 52) and how many extra days are left. In these exact weeks, one Friday will be there for sure (i.e., 52 Fridays) and in the remaining days what are the outcomes which will lead to the occurrence of one Friday in any one of the days remaining in order to get 53 Fridays.