Question

In a non-leap year, what is the probability that the last day of the year starts with a ‘t’?

Answer 1

Hint: The given question revolves around the concepts of probability. We know that a week consists of seven days. We have to find the probability that the last day of a non-leap year starts with a ‘t’. We find the number of favorable outcomes and total number of outcomes and then find the probability of the event.

Complete step-by-step answer:
In the given question, we have to find the probability such that the last day of a non-leap year starts with a ‘t’.
We know that there are a total of seven days in a week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.
The names of the days starting with ‘t’ are: Tuesday and Thursday. So, the last day of the non-leap year must be a Tuesday or a Thursday.
So, there are only two days in a week that start with ‘t’.
So, the total number of favorable outcomes is $ 2 $ .
Number of total outcomes is $ 7 $ as there are seven days in a week.
Now, we know that the probability of an event to happen is $ \dfrac{{Number\,of\,favourable\,outcomes}}{{Number\,of\,total\,outcomes}} $
Therefore, we get the probability of last day of a non-leap year starting with a ‘t’ is $ \dfrac{{Number\,of\,favourable\,outcomes}}{{Number\,of\,total\,outcomes}} = \dfrac{2}{7} $ .
So, the probability that the last day of a non-leap year starts with a ‘t’ is $ \dfrac{2}{7} $ .
So, the correct answer is “ $ \dfrac{2}{7} $ ”.

Note: We know that there are a total of $ 365 $ days in a non-leap year. So, a non-leap year consists of a total of $ 52 $ complete weeks and one odd day. This information is very useful when attempting probability questions related to days and calendars. We must know the basic concepts and formulae of probability to attempt and solve the given problem. We should be careful while carrying out the calculations so as to be sure of the answer.