Question Number 8803 by desniati@rocketmail.com last updated on 28/Oct/16
$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{a}\:\mathrm{leap}\:\mathrm{year}\:\mathrm{selected} \\ $$$$\mathrm{at}\:\mathrm{random}\:\mathrm{contains}\:\mathrm{either}\:\mathrm{53}\:\mathrm{Sundays}\:\mathrm{or} \\ $$$$\mathrm{53}\:\mathrm{Mondays},\:\mathrm{is} \\ $$
Answered by myintkhaing last updated on 02/Nov/16
$${number}\:{of}\:{days}\:{in}\:{leap}\:{year}=\mathrm{366} \\ $$$${number}\:{of}\:{days}\:{in}\:\mathrm{52}\:{weeks}=\mathrm{364} \\ $$$${thes}\:{extra}\:\mathrm{2}\:{days}\:{may}\:{be} \\ $$$$\left({Sun}\:{Mon}\right)\left({Mon}\:{Tue}\right)\left({Tue}\:{Wed}\right) \\ $$$$\left({Wed}\:{Thu}\right)\left({Thu}\:{Fri}\right)\left({Fri}\:{Sat}\right)\left({Sat}\:{Sun}\right) \\ $$$${P}\left(\mathrm{53}^{{rd}} \:{Sundays}\right)={P}\left({A}\right)=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${P}\left(\mathrm{53}^{{rd}} \:{Mondays}\right)={P}\left({B}\right)=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${P}\left({AB}\right)=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$${P}\left({A}\:{or}\:{B}\right)={P}\left({A}\right)+{P}\left({B}\right)−{P}\left({AB}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{7}}+\frac{\mathrm{2}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{3}}{\mathrm{7}} \\ $$