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Question
The probability that a leap year will have 53 Fridays or 53 Saturdays is
Options
\[\frac{2}{7}\]
\[\frac{3}{7}\]
\[\frac{4}{7}\]
\[\frac{1}{7}\]
Solution
For a non-leap year:
52 weeks + 1 day
For a leap year:
52 weeks + 2 days
\[\text{ Sample space } = [ \left( \text{ Monday, Tuesday } \right), \left( \text{ Tuesday, Wednesday } \right), \left( \text{ Wednesday, Thursday } \right), \]
\[\left( \text{ Thursday, Friday } \right), \left( \text{ Friday, Saturday }\right), \left( \text{ Saturday, Sunday }\right), \left( \text{ Sunday, Monday } ) \right]\]
\[\text{ Favourable cases } = 3\]
\[P\left( 53 \text{ Fridays or 53 Saturdays} \right) = \frac{3}{7}\]
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