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Question
The probability that a leap year will have 53 Fridays or 53 Saturdays is
Options
2/7
3/7
4/7
1/7
Solution
3/7
We know that a leap year has 366 days (i.e. 7 \[\times\] 52 + 2) = 52 weeks and 2 extra days
The sample space for these 2 extra days is given below:
S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
There are 7 cases.
∴ n(S) = 7
Let E be the event that the leap year has 53 Fridays or 53 Saturdays.
E = { (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}
i.e. n(E) = 3
\[\therefore P\left( E \right) = \frac{n\left( E \right)}{n\left( S \right)} = \frac{3}{7}\]
Hence, the probability that a leap year has 53 Fridays or 53 Saturdays is \[\frac{3}{7}\] .
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