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Find the probability that a leap year selected at random

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antoxavier
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Find the probability that a leap year selected at random [#permalink] New post   Updated on: 31 Jan 2013, 04:49
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Find the probability that a leap year selected at random will have 53 Sundays

A. 6/7
B. 5/7
C. 4/7
D. 3/7
E. 2/7
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E

Originally posted by antoxavier on 30 Jan 2013, 23:30.
Last edited by Bunuel on 31 Jan 2013, 04:49, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Find the probability that a leap year selected at random [#permalink] New post   31 Jan 2013, 04:48
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antoxavier wrote:
Find the probability that a leap year selected at random will have 53 Sundays

A. 6/7
B. 5/7
C. 4/7
D. 3/7
E. 2/7


There are 366 days in a leap year: 52 weeks and 2 more days. So, 52 Sundays and 2 days.

These 2 days can be: {Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, {Sat, Sun} and {Sun, Mon} (7 cases). In order to have 53 Sundays we should have either {Sat, Sun} or {Sun, Mon} case.

The probability of that is 2/7.

Answer: E.

Hope it's clear.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3 (the name of a topic MUST be the first 40 characters (~the first two sentences) of the question).
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Probability: [#permalink] New post   21 Mar 2017, 21:14
What is the probability that a leap year selected at random will contain 53 Sundays?

A.1/7 B. 2/7 C.7/366 D.3/7 E. 52/366
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Re: Find the probability that a leap year selected at random [#permalink] New post   21 Mar 2017, 21:32
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techiesam wrote:
What is the probability that a leap year selected at random will contain 53 Sundays?

A.1/7 B. 2/7 C.7/366 D.3/7 E. 52/366


Merging topics. Please refer to the solution above.
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Find the probability that a leap year selected at random [#permalink] New post   22 Mar 2017, 02:27
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Any given year will always have 52 complete weeks. So 52 Sundays will always be there in any year.

A leap year has 366 days.
52 weeks = 52*7=364 days

So we have 2 extra days apart from 52 weeks.

These 2 days can be (Saturday, Sunday), (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday,Thursday), (Thursday, Friday), (Friday, Saturday).

For the year to have 53 Sundays, the 2 days must be either (Saturday, Sunday) or (Sunday, Monday)

Favourable cases = 2
Total possible cases = 7

Propability = \(\frac{Favourable}{Total}\)= \(\frac{2}{7}\)

Answer E
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Re: Find the probability that a leap year selected at random [#permalink] New post   15 Sep 2017, 10:51
There are 366 days in a leap year that contain 52 weeks and 2 more days. So, 52 Sundays and 2 days.

These 2 days can be:
(Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, {Sat, Sun} and {Sun, Mon} (7 cases).

In order to have 53 Sundays we should have either {Sat, Sun} or {Sun, Mon} case.
No. of sample spaces = 7.
No. of event that gives 53rd Sunday in Leap Year = 2.

Required Probability = 2⁄7.

Hit Kudos! :thumbup: :-)
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Re: Find the probability that a leap year selected at random [#permalink] New post   15 Jul 2019, 19:16
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Does GMAT ask these type of questions?

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Re: Find the probability that a leap year selected at random [#permalink] New post   10 Sep 2020, 04:34
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A leap year has 366 days.

=> \(\frac{366 }{ 7}\) = 52 weeks and 2 days remainder

There is always 1 Sunday in a week, therefore in 52 weeks, we have 52 Sundays.

The remaining 2 days can be SAT+SUN, SUN+MON, MON+TUE, TUE+WED, WED+THUR, THUR+FRI, FRI+SAT.

=> Total possible outcomes: 7

=> Desired: 2 [SAT+SUN, SUN+MON]

=> Probability: \(\frac{Desired }{ Total}\)

=> \(\frac{2}{7}\)

Answer E
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Re: Find the probability that a leap year selected at random [#permalink] New post   07 Mar 2024, 18:00
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Re: Find the probability that a leap year selected at random [#permalink]
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