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Question Number 8803 by desniati@rocketmail.com last updated on 28/Oct/16

The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays, is

$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{a}\:\mathrm{leap}\:\mathrm{year}\:\mathrm{selected} \\ $$$$\mathrm{at}\:\mathrm{random}\:\mathrm{contains}\:\mathrm{either}\:\mathrm{53}\:\mathrm{Sundays}\:\mathrm{or} \\ $$$$\mathrm{53}\:\mathrm{Mondays},\:\mathrm{is} \\ $$

Answered by myintkhaing last updated on 02/Nov/16

number of days in leap year=366 number of days in 52 weeks=364 thes extra 2 days may be (Sun Mon)(Mon Tue)(Tue Wed) (Wed Thu)(Thu Fri)(Fri Sat)(Sat Sun) P(53^(rd) Sundays)=P(A)=(2/7) P(53^(rd) Mondays)=P(B)=(2/7) P(AB)=(1/7) P(A or B)=P(A)+P(B)−P(AB) =(2/7)+(2/7)−(1/7)=(3/7)

$${number}\:{of}\:{days}\:{in}\:{leap}\:{year}=\mathrm{366} \\ $$$${number}\:{of}\:{days}\:{in}\:\mathrm{52}\:{weeks}=\mathrm{364} \\ $$$${thes}\:{extra}\:\mathrm{2}\:{days}\:{may}\:{be} \\ $$$$\left({Sun}\:{Mon}\right)\left({Mon}\:{Tue}\right)\left({Tue}\:{Wed}\right) \\ $$$$\left({Wed}\:{Thu}\right)\left({Thu}\:{Fri}\right)\left({Fri}\:{Sat}\right)\left({Sat}\:{Sun}\right) \\ $$$${P}\left(\mathrm{53}^{{rd}} \:{Sundays}\right)={P}\left({A}\right)=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${P}\left(\mathrm{53}^{{rd}} \:{Mondays}\right)={P}\left({B}\right)=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${P}\left({AB}\right)=\frac{\mathrm{1}}{\mathrm{7}} \\ $$$${P}\left({A}\:{or}\:{B}\right)={P}\left({A}\right)+{P}\left({B}\right)−{P}\left({AB}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{7}}+\frac{\mathrm{2}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{7}}=\frac{\mathrm{3}}{\mathrm{7}} \\ $$

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