What is the probability that a leap year, selected at random will contain (i) 53 Fridays (ii) 53 Mondays and 53 Tuesdays (iii) 53 Sundays and 53 Thursdays.

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Solution

A leap year has 366 days, which works out to 52 full weeks and two additional days.
These two additional days could be:
Sunday - Monday
Monday - Tuesday
Tuesday - Wednesday
Wednesday - Thursday
Thursday - Friday
Friday - Saturday
Saturday - Sunday.

These seven outcomes are equally likely.
(i) Thus, for having 53 mondays from last two days one day should be friday i.e.
Friday - Saturday
Thursday - Friday
Thus the probability that one of these two days is Friday is
$= \frac{F a v o u r a b l e o u t c o m e s}{T o t a l o u t c o m e s}$
$= \frac{2}{7}$ .
(ii) There’s only one outcome above which accounts for 53 Mondays and 53 Tuesdays i.e Monday - Tuesday . Therefore, the probability is
$= \frac{F a v o u r a b l e o u t c o m e s}{T o t a l o u t c o m e s}$
$= \frac{1}{7}$ .
(iii) There is no outcome above which will account for 53 Sundays and 53 Thursdays. So the probability is 0 (imposible event).

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