Let \( A \) be the event that at least two people share the same week day birthday.
Then,
\[
P(A) = 1 - P(\text{all birthdays fall on different week days})
\]
When \( k = 3 \), the number of ways to assign different week days is:
\[
P(\text{all distinct} \mid k = 3)=
\frac{7}{7} \cdot \frac{6}{7} \cdot \frac{5}{7} = \frac{210}{343} = \frac{30}{49}
\]
When \( k = 5 \), the number of ways to assign different week days is:
\[P(\text{all distinct} \mid k = 5)=
\frac{7}{7} \cdot \frac{6}{7} \cdot \frac{5}{7} \cdot \frac{4}{7} \cdot \frac{3}{7} = \frac{2520}{16807} = \frac{360}{49^2}
\]
When \( k = 10 \), we are assigning 10 people, different week days which is not possible (by PHP)
\[
P(\text{all distinct} \mid k = 10) = 0
\]
Given:
\[
P(k = 3) = \frac{1}{4}, \quad P(k = 5) = \frac{1}{4}, \quad P(k = 10) = \frac{1}{2}
\]
Applying total probability:
\[
P(\text{all distinct}) = \frac{1}{4} \cdot \frac{30}{49} + \frac{1}{4} \cdot \frac{360}{49^2} + \frac{1}{2} \cdot 0
\]
\[
= \frac{1}{4} \left( \frac{30}{49} + \frac{360}{49^2} \right)
\]
\[
= \frac{15}{98} \left( 1 + \frac{12}{49} \right)
\]
Thus, the required probability is
\[
P(A) = 1 - \frac{15}{98} \left( 1 + \frac{12}{49} \right)
\]
