find the probability that a leap year selected at random will contain 53 Sundays?
Answers
We know that a leap year contains 366 days and these days are as 52 weeks and 2 odd days.
Days which remain not as a week, just like the remainder on dividing a number by another, is called odd day.
Since a leap year contains 52 weeks and 2 odd days, we can say that a leap year selected at random may contain two consecutive days 53 times - 52 such days are taken in the 52 weeks and the remaining one, each, remain as odd days.
These two odd days can be of which days? Well, they can be either of the following possibilities.
(1) Sunday and Monday
(2) Monday and Tuesday
(3) Tuesday and Wednesday
(4) Wednesday and Thursday
(5) Thursday and Friday
(6) Friday and Saturday
(7) Saturday and Sunday
So these are the total outcomes. Among these, there are two possible outcomes in which Sunday occurs.
(1) Sunday and Monday
This implies the leap year should start on Sunday, since each week ends on Saturday. Here the year contains 53 Sundays as well as Mondays.
(2) Saturday and Sunday
This implies the leap year should start on Saturday, since each week ends on Friday. Here the year contains 53 Saturdays as well as Sundays.
So the probability that the leap year contains 53 Sundays is equal to 2 / 7.
Answer:
In a leap year there are 366 days. In 366 days, we have 522 weeks and 2 days, Thus we can say that leap year ah always 52 Sundays.
The remaining two days can be :
(i) Sunday and Mondays
(ii) Mondays and Tuesdays
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturady and Sunday.
From above it is clear that there are 7 elementary events associated with this random experiment.
Clearly the event A will happen if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday.