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Question
In a non-leap year, the probability of having \(53\) Tuesdays or \(53\) Wednesdays is
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Solution
There are \(365\) days in a non-leap year.
\(365\) days \(= 52\) weeks \(+ 1\) day
Which means, every day of the week will come at least \(52\) times in a year.
To have \(53\) Tuesdays or \(53\) Wednesdays in a year, the remaining one day should be Tuesday or Wednesday.
Since, the remaining one day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday.
Total number of cases, \(n(S)=7\)
Since, to have \(53\) Tuesdays or \(53\)
Wednesdays in a year, the remaining one day should be Tuesday or Wednesday.
\(\therefore\) No. of favorable cases, \(n(E)=2\)
Required probability
\(=\dfrac{Number ~offavorable ~cases}{total~ number ~of ~cases }\)
\(=\dfrac{2}{7}\)
Final Answer: B
\(365\) days \(= 52\) weeks \(+ 1\) day
Which means, every day of the week will come at least \(52\) times in a year.
To have \(53\) Tuesdays or \(53\) Wednesdays in a year, the remaining one day should be Tuesday or Wednesday.
Since, the remaining one day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday or Saturday.
Total number of cases, \(n(S)=7\)
Since, to have \(53\) Tuesdays or \(53\)
Wednesdays in a year, the remaining one day should be Tuesday or Wednesday.
\(\therefore\) No. of favorable cases, \(n(E)=2\)
Required probability
\(=\dfrac{Number ~offavorable ~cases}{total~ number ~of ~cases }\)
\(=\dfrac{2}{7}\)
Final Answer: B
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