Question

Find the probability that a leap year selected at random has 53 sundays

Answer 1

No. of days in a leap year = 366 days
No. of weeks = 366/ 7 = 52 weeks and 2 days

So for 53 Sundays one of the last 2 days should be sunday.
Sample space  : Saturday , Sunday
                         Sunday , Monday
                        Monday , Tuesday
                           Tuesday , Wednesday
                       Wednesday , Thursday
                       Thursday, Friday
                          Friday , Saturday

And we require (Saturday , Sunday) and (Sunday , Monday)

So prob. = 2/7

Hope this helps.

Answer 2

Answer:

In a leap year there are 366 days. In 366 days, we have 522 weeks and 2 days, Thus we can say that leap year ah always 52 Sundays.

The remaining two days can be

(i) Sunday and Mondays

(ii) Mondays and Tuesdays

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(VI) Friday and Saturday

(vii) Saturady and Sunday.

From above it is clear that there are 7 elementary events associated with this random experiment.

Clearly the event A will happen if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday.

∴ P (E) = n (E)/n (S) = 2/7