Question

In a non leap year find the probability of 53 Mondays

Answer 1

As we know that : There are 365 days in non leap year.

$7 \: weeks + 1 \: day$

So, there are 52 Monday in 7 weeks.

Now we have to find the Probability of 53 Mondays. For that the extra 1 day should have Monday.

So, the Probability of extra 1 day to be Monday

As we know that :

$Probability = \frac{Number \: of \: favourable \: events}{Number \: of \: total \: evets}$

$Now ,\: favourable \: event \: E= Monday \\ \\ = > Number \: of \: favourable\: events n(E) = 1 \\ \\ and \\ \\ total \: events \: S = Sunday, \: Monday,....... Saturday \\ \\ = > Number \: of \: total \: events \: n(S) = 7 \\ \\ now \: Probability \: P = \frac{n(E)}{n(S)} \\ \\ = > P = \frac{1}{7}$

So, the probability of 53 Mondays in non leap year will be 1/7