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(D) Let $n$ be chosen from $\{1, 2, \ldots, 100\}$ and the starting day $d$ be chosen from $\{1, 2, \ldots, 7\}$ (where $1$ is Wednesday,$2$ is Thursd

Vedclass · Accepted Answer

$\frac{43}{175}$

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(D) Let $n$ be chosen from $\{1, 2, \ldots, 100\}$ and the starting day $d$ be chosen from $\{1, 2, \ldots, 7\}$ (where $1$ is Wednesday,$2$ is Thursday,...,$7$ is Tuesday). Let $S(n, d)$ and $M(n, d)$ be the number of Sundays and Mondays in the $n$ consecutive days starting from day $d$. We want to find the probability that $S(n, d) 
eq M(n, d)$. For any $n$,we can write $n = 7q + r$,where $0 \leq r In any sequence of $7q$ days,there are exactly $q$ Sundays and $q$ Mondays. Thus,$S(n, d) = q + S(r, d)$ and $M(n, d) = q + M(r, d)$. The condition $S(n, d) 
eq M(n, d)$ is equivalent to $S(r, d) 
eq M(r, d)$. This occurs if and only if the remainder $r$ contains a Sunday but not a Monday,or a Monday but not a Sunday. For a fixed $r \in \{1, 2, \ldots, 6\}$,there are $7$ possible starting days. By analyzing the cycles,for each $r$,there are exactly $2$ starting days out of $7$ where the number of Sundays and Mondays differ. Thus,for any $n$ such that $n \pmod{7} 
eq 0$,the probability is $\frac{2}{7}$. If $n \pmod{7} = 0$,the probability is $0$. There are $14$ values of $n$ in $\{1, \ldots, 100\}$ such that $n$ is a multiple of $7$ $(7, 14, \ldots, 98)$. There are $86$ values of $n$ such that $n \pmod{7} 
eq 0$. The total probability is $\frac{86}{100} 	imes \frac{2}{7} + \frac{14}{100} 	imes 0 = \frac{172}{700} = \frac{43}{175}$.

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