Question

What is the probability that an ordinary year has 53 Sundays?

Answer 1

SOLUTION :  

Given : An ordinary year.

Total number of days in ordinary year = 365 days .It  contain 52 weeks and 1 day

This one day can be any day of the week :  

Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.

Here, we have to make 53 Sundays so one  additional day should be Sunday.

Total number of days = 7  

Total number of outcomes = 7  

Let E = Event of getting an ordinary year which has 53 Sundays

Number of favourable outcomes : 1 (Sunday)

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 1/7  

Hence, Probability of getting an ordinary year which has 53 Sundays, P(E) = 1/7 .

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Answer 2

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The answer goes here....

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》To find -

Probability that an ordinary year has 53 days ?

》Solution -

Given that an ordinary year is there. Since, we know that an ordinary year has 52 weeks.

But that 52 weeks counts for only 364 days. Also here it is given, that the year is ordinary year which means it has 365 days.

So, it is possible that 53 sundays can be there if the rest of one day is Sunday.

i.e.,

⇒ $Probability$ = $\frac{Favourable\:outcome}{Total\:outcome}$

Here, the favourable outcome is sunday which is 1.

⇒ $Probability$ = $\frac{1}{7}$

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