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If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

$\begin{array}{1 1} \frac{2}{7} \\ \frac{1}{7} \\ \frac{2}{53} \\ \frac{1}{53} \end{array} $

1 Answer

There are 7 days in a week x 52 weeks = 364 days. A leap year has 366 days, so it must have 52 weeks and two extra days.
The 52 weeks have 52 tuesdays obviously, so we need to calculate the probability that one of the two extra days is a Tuesday.
The sample space for those 2 extra days is as follows: $\begin{Bmatrix} S, M & M, T & T, W & W, Th & Th, F & F, S\\ \end{Bmatrix}$
The favourable outcomes for us are $\begin{Bmatrix} M, T & T, W \\ \end{Bmatrix}$.
Therefore P (extra Tuesday) = $\large\frac{2}{7}$
answered Jun 22, 2013 by balaji.thirumalai

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