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Answers
Answer:
2/7
Step-by-step explanation:
Number of days in a leap year = 366. It contains 52 weeks + 2 days.
The remaining 2 days can be one of the following seven cases:
(i) Sunday,Monday
(ii) Monday,Tuesday
(iii) Tuesday,Wednesday
(iv) Wednesday,Thursday
(v) Thursday,Friday
(vi) Friday,Saturday
(vii) Saturday, Sunday
Total number of possible cases n(S) = 7.
Now,
For a leap year to contain 53 Mondays,last two days are either Sunday and Monday (or) Saturday and Sunday.
∴ Number of such favorable cases n(A) = 2
Therefore, Probability of getting 53 Mondays = n(A)/n(S)
= 2/7.
Hope it helps!
A normal year has 52 Mondays, 52 Tuesdays, 52 Wednesdays, 52 Thursdays, 52 Fridays, 52 Saturdays and 52 Sundays + 1 day that could be anything depending upon the year under consideration. In addition to this, a leap year has an extra day which might be a Monday or Tuesday or Wednesday...or Sunday.
We've now reduced the question to : what is the probability that in a given pair of consecutive days of the year one of them is a Sunday?
Our sample space is S : {Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday,..., Sunday-Monday}
Number of elements in S = n(S) = 7
What we want is a set A that comprises of the elements Saturday-Sunday and Sunday-Monday i.e. A : {Saturday-Sunday, Sunday-Monday}
Number of elements in set A = n(A) = 2
By definition, probability of occurrence of A = n(A)/n(S) = 2/7
Therefore, probability that a leap year has 53 Sundays is 2/7. (Note that this is true for any day of the week, not just Sunday)
I think it helps you