Question

find the probability of getting 53 Fridays in a non leap year and a leap year

Answer 1

(1) 53 Fridays in a leap year:

We know that a leap year consists of 366 days. 52 weeks + 2 days.

These 2 days can be:

(i) {Mon, Tue}

(ii) {Tues, Wed}

(iii) {Wed, Thurs}

(iv) {Thurs,Fri}

(v) {Fri, Sat}

(vi) {Sat,Sun}

(vii) {Sun,Mon}

Total number of favorable cases n(S) = 7.

Now,

Number of cases we get Fridays n(A) = {Thurs,fri},{fri,sat} = 2

Required probability P(A) = n(A)/n(S) = 2/7.

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(2)

We know that A non leap year has 365 days.

A year has 52 weeks.

A week has 7 days.

= > 52 * 7 = 364.

So, 365 - 364 = 1.

That means this 1 days can be = sun, mon,tues,wed,thurs,fri,sat.

Number of cases = 7.

Now,

Probability of getting 53 fridays = 1.

Therefore, required probability P(A) = 1/7.

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Hope it helps!

Answer 2

Leap year:- The year ( every four years ) with 366 days instead of 365 is called leap year.