Find the probability that a leap year, selected at random, will contain 53 Sundays.
A. 5⁄7
B. 4⁄7
C. 3⁄7
D. 2⁄7
Answer: Option D
Solution(By Examveda Team)
There are 366 days in a leap year that contain 52 weeks and 2 more days. So, 52 Sundays and 2 days.These 2 days can be:
(Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, {Sat, Sun} and {Sun, Mon} (7 cases).
In order to have 53 Sundays we should have either {Sat, Sun} or {Sun, Mon} case.
No. of sample spaces = 7.
No. of event that gives 53rd Sunday in Leap Year = 2.
Required Probability = 2⁄7.
This Question Belongs to User Ask Question >> Miscellaneous
Join The Discussion
Comments ( 2 )
Related User Ask Questions
Which of the following is not a primary function of a Bank?
A. Granting Loans
B. Collecting Cheques/Drafts customers
C. Facilitating import of goods
D. Issuing Bank Drafts
A. Regulatory jurisdictional fight between SEBI and IRDA
B. They don’t offer better tax benefits
C. They offer lesser returns compared to traditional insurance policies
D. All of the above
The Chameli Devi Jain Award is given for an outstanding woman ____?
A. Scientist
B. Reporter
C. Player
D. Teacher
Thanx it helps me a lot
In a leap year, there are 366 days and 364days make 52 weeks and in each week there is 1 Sunday. Therefore, we have to find the probability of having 1 Sunday out of the remaining 2 days.
Now, the 2 days can be (Sunday,Monday) or (Monday,Tuesday) or (Tuesday, Wednesday) or (Wednesday, Thursday) or (Thursday,Friday) or (Friday,Saturday) or (Saturday,Sunday). Note that Sunday occurs 2 times in these 7 pairs.
Let the event be 'having a Sunday', then the number of favourable outcomes to the event = 2
Therefore, Probability(having 53 Sundays) = 2/7.