Question

The probability of 53 Sundays in a non-leap year is

A. 1/7

B. 2/7

C. 53/365

D. 53/366​

Answer 1

TO FIND :–

The probability of 53 Sundays in a non-leap year = ?

SOLUTION :–

• We know that –

=> Total days in a non-leap year = 365

• 365 days = 52 weeks + 1 day

• 52 week means 52 sundays.

• For 53 sunday => one remaining should be Sunday.

→ E = Sunday

→ n(E) = 1

• And –

→ S = { Sunday, Monday, Tuesday, Wednesday , Thursday, Friday , Saturday}

→ n(S) = 7

➣ Probability = n(E)/n(S)

• So that –

⇒ Probability = 1/7

▪︎ Hence , Option (A) is correct.

Answer 2

Given:

• To find the probability of 53 Sundays in a non leap year.

Answer:

As we know that a non- leap year has 365 days .

Now , So total number of weeks in a leap year = (365/7) , where the number of weeks is equal to the whole number part of the quotient.

And ( 365/7) = 52.14 which means a non - leap year has 52 weeks .

No. of days left = ( 365-52×7) = 365-364 = 1.

Hence this one day could be Sunday , Monday, Tuesday, Wednesday,Thursday,Friday or Saturday.

Here ,

• Total number of possible outcomes = 7
• Total number of favourable outcomes = 1.

Therefore P(of getting a Sunday) = Total Number of favourable outcomes/Total Number of outcomes = 1/7.

Hence the required answer is 1/7 (a)