Question

What is the probability that a leap year , selected at random will contain 53 Sundays ?

Answer 1

 A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days. 
The remaining 2 days may be any of the following : 
(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 

For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday. 

n(S) = 7 
n(E) = 2 (As two of the seven options contains Sunday)
P(E) = n(E) / n(S) = 2 / 7

Answer 2

In a leap year there are 366 days, right ........ so it makes ( 52 weeks and 2 'extra' days.) So on the 53rd Sunday can occur in one of these 2 days. P(E) = n(E) / n(S) = 2 / 7 Therefore the probability is 2/7