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Question
What is the probability that a non-leap year has 53 Sundays?
Options
\[\frac{6}{7}\]
\[\frac{1}{7}\]
\[\frac{5}{7}\]
None of these
Solution
A non leap year
TO FIND: Probability that a non leap year has 53 Sundays.
Total number of days in a non leap year is 365days
Hence number of weeks in a non leap year is `365/7=52 "Weeks and 1 day"`
In a non leap year we have 52 complete weeks and 1 day which can be any day of the week i.e. Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday
To make 53 Sundays the additional day should be Sunday
Hence total number of days which can be any day is7
Favorable day i.e. Sunday is 1
`"We know that PROBABILITY" ="Number of favourable event"/"Total number of event"`
Hence probability that a non leap year has 53 Sundays is `1/7`
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