Question

find the probablity that a leap year selected at random will contain 53sunday

Answer 1

Answer:

it is 2 / 7

i guess

A leap year has 366 days or 52 weeks and 2 odd days.

The two odd days can be

Sunday,Monday

Monday,Tuesday

Tuesday,Wednesday

Wednesday,Thursday

Thursday,Friday

Friday,Saturday

Saturday,Sunday

So there are 7 possibilities out of which 2 have a Monday. So

the probability of 53 Mondays is 2/7.

Answer 2

AnswEr:

In a leap year there are 366 days

• 366 days = 52 weeks and 2 days

Thus, a leap year has always 52 Sundays. The remaining 2 days can be :

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

If S is the sample space associated with the experiment, then S consists of the above seven points :

$\therefore$$\sf{Total\:numbers\:of\:elementary\:events=7}$

Let A be the event that a leap year has 53 Sundays. In order that a leap year, selected at random, should contain 53 Sundays, one of the 'over' days must be a Sunday. This can be in any one of the following ways :

1. Sunday and Monday
2. Saturday and Sunday

$\therefore$ $\sf{Favourable\:numbers\:of\:elementary\:outcomes}$

= 2

Hence, required probability = 2/7.