Question

Find the probability that a common year (not leap year) contains only 52 Sundays

Answer 1

I think your quest is wrong . It should be 53 Sundays. Do recheck.
I am giving the solution of the common year having 53 Sundays:
1 year=365 days
1 week = 7 days
365/7=52weeks +1 day
Let 'S' be the Samples Space of this 1 day
S={Sun,Mon,Tues,Wed,Thurs,Fri,Sat}
n(S)=7
Let 'A' be the event of this 1 day to be Sunday
A={Sun}
n(A)=1
P(A)=n(A)/n(S)=1/7

#Do recheck the question