Question

Find out the probability of getting 53 fridays and 53 saturdays in a leap year?

Answer 1

$$\begin{gathered} \mathrm{Dear}\mathrm{Student}, \\ \mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{days}\mathrm{in}\mathrm{a}\mathrm{leap}\mathrm{year}=366 \\ \mathrm{Now},366\mathrm{days}=52\mathrm{weeks}+2\mathrm{days} \\ \mathrm{So},\mathrm{there}\mathrm{are}7\mathrm{possible}\mathrm{combinations}\mathrm{for}\mathrm{the}\mathrm{last}2\mathrm{days}\mathrm{of}\mathrm{a}\mathrm{leap}\mathrm{year}: \\ 1.\mathrm{Sunday},\mathrm{Monday} \\ 2.\mathrm{Monday},\mathrm{Tuesday} \\ 3.\mathrm{Tuesday},\mathrm{Wednesday} \\ 4.\mathrm{Wednesday},\mathrm{Thursday} \\ 5.\mathrm{Thursday},\mathrm{Friday} \\ 6.\mathrm{Friday},\mathrm{Saturday} \\ 7.\mathrm{Saturday},\mathrm{Sunday} \\ \mathrm{Out}\mathrm{of}\mathrm{these}7,\mathrm{only}1\mathrm{combination}\mathrm{gives}\mathrm{us}53\mathrm{fridays}\mathrm{and}53\mathrm{saturdays}. \\ \mathrm{Required}\mathrm{probability}=\frac{1}{7} \\ \mathrm{Regards} \end{gathered}$$