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The probability of a non-leap year having 53 Mondays is .........
Options:
Solution:
2807 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{7}$
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.
$$
\begin{aligned}
& =52 \times 7+1 \\
& =52 \text { weeks }+1 \text { day }
\end{aligned}
$$
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.
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