find the probability that a leap year selected at random will contain 53 sundays
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Answered by
977
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.
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 (say) 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)
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 (say) 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)
Answered by
861
A leap year has 366 days or 52 weeks and 2 odd days. The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday},{Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}. So there are 7 possibilities out of which 2 have a Sunday. So the probability of 53 Sundays is 2/7.
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