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# Find the probability that a leap year selected at random

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Intern
Joined: 25 Jan 2013
Posts: 24
Find the probability that a leap year selected at random  [#permalink]

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Updated on: 31 Jan 2013, 04:49
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Difficulty:

35% (medium)

Question Stats:

68% (00:51) correct 32% (01:09) wrong based on 212 sessions

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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

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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Posts: 47967
Re: Find the probability that a leap year selected at random  [#permalink]

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31 Jan 2013, 04:48
6
6
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.

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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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]

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21 Mar 2017, 21:32
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]

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22 Mar 2017, 02:27
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 = $$\frac{Favourable}{Total}$$= $$\frac{2}{7}$$

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Re: Find the probability that a leap year selected at random  [#permalink]

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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!
Re: Find the probability that a leap year selected at random &nbs [#permalink] 15 Sep 2017, 10:51
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