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Find the probability that a randomly selected year has 53 Sundays.

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Find the probability that a randomly selected year has 53 Sundays.  [#permalink]

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New post 30 Jan 2019, 22:23
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Find the probability that a randomly selected year has 53 Sundays.
A) \(\frac{1}{7}\)
B) \(\frac{2}{7}\)
C) \(\frac{3}{7}\)
D) \(\frac{3}{28}\)
E) \(\frac{5}{28}\)

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Re: Find the probability that a randomly selected year has 53 Sundays.  [#permalink]

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New post 31 Jan 2019, 08:00
4d wrote:
Find the probability that a randomly selected year has 53 Sundays.
A) \(\frac{1}{7}\)
B) \(\frac{2}{7}\)
C) \(\frac{3}{7}\)
D) \(\frac{3}{28}\)
E) \(\frac{5}{28}\)



The question involves knowing leap year and number of days in a year..

So, let us talk of this question in two different cases..
(1) A normal year
It has 365 days which is 52 weeks and 1 day. So any of the 7 days from Sunday to Saturday that the year starts with will be 53 and rest 6 will be 52. Probability of Sunday will be as any of other seven so \(\frac{1}{7}\).
(2) A Leap year
It has 366 days which is 52 weeks and 2 days. So any 2 of the 7 days from Sunday to Saturday that the year starts with will be 53 and rest 5 will be 52. Probability of Sunday will be as any of other seven so \(2*\frac{1}{7}=\frac{2}{7}\). Multiplication by 2 is because we have 2 days that can be 53.

Now we have to check the probability of a normal year and a leap year.
Here, we will take a standard case that every 4th year is leap year and will not consider the 1000th year as non-leap
so Probability of normal year is 3/4 and that of leap year is 1/4.

Answer \(\frac{3}{4}*\frac{1}{7}+\frac{1}{4}*\frac{2}{7}=\frac{5}{28}\)

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Re: Find the probability that a randomly selected year has 53 Sundays.  [#permalink]

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New post 31 Jan 2019, 09:25
4d wrote:
Find the probability that a randomly selected year has 53 Sundays.
A) \(\frac{1}{7}\)
B) \(\frac{2}{7}\)
C) \(\frac{3}{7}\)
D) \(\frac{3}{28}\)
E) \(\frac{5}{28}\)


Total days in year = 365 or say 52 weeks and 1 day
Total days in leap year = 366 days or say 52 weeks and 2 days

so P of sunday in non leap year = 1/7
and P of sunday in leap year = 2*1/7 ; reason why since we have a day extra in a leap year so chances are that sunday can be on that extra day

P of a leap year is 1/4 and non leap year is 3/4
so for sunday
1/4*2/7 + 3/4*1/7
2+3/28 ; 5/28
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Find the probability that a randomly selected year has 53 Sundays.  [#permalink]

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New post 11 Feb 2019, 20:24
Non leap year case: (365 % 7) = 1, (every day in a week, repeats 52 times), now for 53 sundays, the remaining last day has to be sunday, then Jan 1 of that year must be sunday.

probability of 53 sundays = (3/4)(probability of non-leap year) * (1/7) ( probability of jan1 of that year to be sunday) = 3/28

Leap year case: (366 % 7) = 2, (every day in a week, repeats 52 times), now two days remaining, for 53 sundays, either of last two days must be sunday, then Jan 1 of that year has to be either saturday or sunday.

probability of 53 sundays = (1/4) (probability of leap year) * (2/7) (probability of jan 1 of that year to be either sunday or saturday) = 2/28

so total probability (leap year case + non leap year case) = (3/28) + (2/28) = 5/28
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Find the probability that a randomly selected year has 53 Sundays.   [#permalink] 11 Feb 2019, 20:24
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