Question

What is the probability that a non-leap year has 53 Sundays?
(a)$\frac{6}{7}$
(b)$\frac{1}{7}$
(c)$\frac{5}{7}$
(d)None of these

Answer 1

SOLUTION :  

The correct option is (b) : 1/7

Given : A non leap year (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 a non leap 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 a non leap year which has 53 Sundays, P(E) = 1/7 .

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Answer 2

$\textsf{Answer :}$

Option B is correct.

$\textsf{Step-by-step explanation :}$

Given that a non-leap year is there. Since, we know that a non-leap year has 52 weeks.

But that 52 weeks counts for only 364 days. Also here it is given, that the year is non-leap 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.

Now, the possibe outcomes are Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday. Here, we get Sunday once.

So, the number of favourable outcome is 1 and that of total outcome is 7.

So, probability -

$\large {\implies \frac{Number\:of\:favourable\:outcomes}{Total\:number\:of\:outcomes}}$

$\large{\implies \frac{1}{7}}$

Thus, option B is correct.