Question

what is the probability that a leap year selected at random will contain 53 thursdays

Answer 1

the answer is 53/366

Answer 2

$no. \: of \: total \: outcomes = 7(mon. \: nd \: tue. \: \: tue. \: nd \: wed. \: \: wed. \: nd \: thur. \: \: thur. \: nd \: fri. \: \: fri. \: nd \: sat. \: \: sat. \: nd \: sun. \: \: sun. \: nd \: mon.)$
$no. \: of \: favourable \: outcomes = 2(wed. \: nd \: thur. \: \: thur. \: nd \: fri.)$
$p(e) = \frac{2}{7}$