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  4. In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is

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Q. In a non-leap year, the probability of having $53$ Tuesdays or $53$ Wednesdays is

Probability

Solution:

In a non-leap year, there are $365$ days in which $52$ weeks and one day extra.
It may be Mon, Tue, Wed, Thu, Fri, Sat, Sun.
$\therefore $ Required probability
$ = P(53$ Tuesday) $+ P(53$ Wednesday)
$- P(53$ Tuesday $\cap\, 53 $ Wednesday)
$=\frac{1}{7}+\frac{1}{7}-0 $
$= \frac{2}{7}$