jmc

There are four possible cases for the Cubs winning the World Series, depending on the number of games that Red Sox win before the Cubs win their fourth game: the Red Sox can win no games, one game, two games, or three games. In general, if the Red Sox win exactly $k$ games before the Cubs win their 4th game, there will be a total of $3+k$ games played before the last one (which the Cubs must win), there will be a total of $\left(\genfrac{}{}{0ex}{}{3+k}{k}\right)$ ways of selecting the games that the Red Sox win out of those, and for each of those arrangements the Cubs will win their 4 games with probability ${\left(\frac{3}{5}\right)}^4$ and the Red Sox will win the $k$ games that are selected for them with probability ${\left(\frac{2}{5}\right)}^k$, so we are left to evaluate the expression $\left(\genfrac{}{}{0ex}{}{3+k}{k}\right){\left(\frac{3}{5}\right)}^4{\left(\frac{2}{5}\right)}^k$ for $k=0,1,2,3$. This gives us our final probability of $$\begin{array}{rl}& \left(\genfrac{}{}{0ex}{}{3}{0}\right){\left(\frac{3}{5}\right)}^4{\left(\frac{2}{5}\right)}^0+\left(\genfrac{}{}{0ex}{}{3+1}{1}\right){\left(\frac{3}{5}\right)}^4{\left(\frac{2}{5}\right)}^1+\\ & \left(\genfrac{}{}{0ex}{}{3+2}{2}\right){\left(\frac{3}{5}\right)}^4{\left(\frac{2}{5}\right)}^2+\left(\genfrac{}{}{0ex}{}{3+3}{3}\right){\left(\frac{3}{5}\right)}^4{\left(\frac{2}{5}\right)}^3\end{array}$$ which simplifies to $$\begin{array}{rl}& 1\cdot (.1296)\cdot 1+4\cdot (.1296)\cdot (.4)\\ & +10\cdot (.1296)\cdot (.16)+20\cdot (.1296)\cdot (.064)=.7102\dots ,\end{array}$$ so our answer is $\boxed{71}$ percent.