The probability of a non-leap year having 53 Mondays is .........
The probability of a non-leap year having 53 Mondays is .........
$\frac{2}{7}$
$\frac{1}{7}$
$\frac{5}{7}$
$\frac{6}{7}$
Solution
Number of days in non Leap year $=365$
$
\begin{aligned}
& =52 \times 7+1 \\
& =52 \text { weeks }+1 \text { day }
\end{aligned}
$
The one additional day may be any one of the day
$
\begin{aligned}
\therefore \quad n(A) & =1 \\
n(s) & =7 \\
\mathrm{P}(\text { Getting } 53 \text { sundays }) & =\frac{n(A)}{n(S)}=\frac{1}{7}
\end{aligned}
$
Hence, option (2) is correct