MCQ
The probability that a leap year selected randomly will have $53$ Sundays is
- A$\frac{1}{7}$
- ✓$\frac{2}{7}$
- C$\frac{4}{{53}}$
- D$\frac{4}{{49}}$
Hence required probability $ = \frac{2}{7}.$
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$\frac{3}{{1! + 2! + 3!}} + \frac{4}{{2! + 3! + 4!}} + \frac{5}{{3! + 4! + 5!}} + ...... + \frac{{2008}}{{\left( {2006} \right)! + \left( {2007} \right)! + \left( {2008} \right)!}}$ is equal to