2007-I-15

$\begin{array}{cl}
= & \dfrac{48}{80} \\
= & \dfrac{3}{5}
\end{array}$

$\begin{array}{cl}
= & \dfrac{12}{80} \\
= & \dfrac{3}{20}
\end{array}$

$\begin{array}{cl}
= & \dfrac{3}{5} + \dfrac{16}{80} – \dfrac{3}{20} \\
= & \dfrac{13}{20}
\end{array}$

$\begin{array}{cl}
= & \dfrac{\frac{3}{20}}{\frac{3}{5}} \\
= & \dfrac{1}{4}
\end{array}$

$\begin{array}{cl}
= & \dfrac{16}{80} \times \dfrac{15}{79} \\
= & \dfrac{3}{79}
\end{array}$

$\begin{array}{cl}
= & \dfrac{28}{80} \times \dfrac{27}{79} + \dfrac{36}{80} \times \dfrac{35}{79} + \dfrac{3}{79} \\
= & \dfrac{141}{395} \\
< & \dfrac{1}{2} \end{array}$

Therefore, the probability of dressing shirts of the same size by the two selected students is not greater than that of dressing different sizes.