The probability that a leap year will have 53 Fridays or 53 Saturdays is
$\dfrac{1}{7}$
$\dfrac{2}{7}$
$\dfrac{3}{7}$
$\dfrac{4}{7}$
Explanation:
There is 366 days In leap year
Now
52 weeks * 7 days/week = 364 day
52 full weeks means -> 52 Friday
For remaining 2 days, one of the following pair is possible
(sun,Mon) (Mon,Tue) (Tue,Wed) (Wed,Thu) (Thu,Fri) (Fri,Sat) (Sat,Sun)
Out of 7 pair there is 2 pair in which 53rd Friday may come.
Probability of 53 Friday in a leap year = (Thu,Fri) (Fri,Sat) = 2/7
Out of 7 pair there is 2 pair in which 53rd Saturday may come.
Probability of 53 Saturday in a leap year = (Fri,Sat) (Sat,Sun) = 2/7
There is 366 days In leap year
Now
52 weeks * 7 days/week = 364 day
52 full weeks means -> 52 Friday
For remaining 2 days, one of the following pair is possible
(sun,Mon) (Mon,Tue) (Tue,Wed) (Wed,Thu) (Thu,Fri) (Fri,Sat) (Sat,Sun)
Out of 7 pair there is 2 pair in which 53rd Friday may come.
Probability of 53 Friday in a leap year = (Thu,Fri) (Fri,Sat) = 2/7
Out of 7 pair there is 2 pair in which 53rd Saturday may come.
Probability of 53 Saturday in a leap year = (Fri,Sat) (Sat,Sun) = 2/7
$p(AUB)=p(A)+p(B)-p(AnB)$
$\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{1}{7}=\dfrac{3}{7}$