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What is the probability of having 53 Sundays or 53 Mondays in a leap year?
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The correct answer is:
$3 / 7$
A leap year has 366 days, in which 2 days may be any one of the following pairs. (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday) (Friday, Saturday) (Saturday, Sunday).
$\therefore \quad$ Required probability $=\frac{2}{7}+\frac{2}{7}-\frac{1}{7}=\frac{3}{7}$
(By using $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\therefore \quad$ Required probability $=\frac{2}{7}+\frac{2}{7}-\frac{1}{7}=\frac{3}{7}$
(By using $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
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