Math, asked by akshitsehgal, 1 year ago

what is the probability that a leap year selected at random will contain 53 Sundays plss explain it prefectly​

Answers

Answered by Anonymous
2

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 (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 add75
2

i hope this helps you

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