what is the probability that a leap year has 53 Sunday
Answers
Answered by
22
A leap yr has 366 days
366 days=52 weeks+2 days extra
S={SM,MT,TW,WT,TF,FS,SS}
n(S)=7
Let A be the event that 2 days contain sunday
A={SM,SS}
n(A)=2
P(A)=n(A)/n(S)
=2/7
366 days=52 weeks+2 days extra
S={SM,MT,TW,WT,TF,FS,SS}
n(S)=7
Let A be the event that 2 days contain sunday
A={SM,SS}
n(A)=2
P(A)=n(A)/n(S)
=2/7
Answered by
13
(1) A non-leap year consists of 365 days.
The number of days in a week n(S) = 7.
365 days = 52 * 7
= 364.
That means we are left with 1-day.
And that 1 day can be Sun, Mon, Tue, Wed, Thu, Fri, Sat.
Therefore n(E) = 1.
(2)A leap year consists of 366 days.
The number of days in a week n(S) = 7.
366 days = 52 * 7
= 364.
That means we are left with 2-days.
Therefore n(E) = 2.
Therefore the probability = n(E)/n(S)
= 2/7.
Hope this helps!
The number of days in a week n(S) = 7.
365 days = 52 * 7
= 364.
That means we are left with 1-day.
And that 1 day can be Sun, Mon, Tue, Wed, Thu, Fri, Sat.
Therefore n(E) = 1.
(2)A leap year consists of 366 days.
The number of days in a week n(S) = 7.
366 days = 52 * 7
= 364.
That means we are left with 2-days.
Therefore n(E) = 2.
Therefore the probability = n(E)/n(S)
= 2/7.
Hope this helps!
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