Let A$=\{$a$,$ b$\}.$ Construct an automaton M which will precisely accept

those words over A which end in two b$$s$.[$corresponding Language L$=$L$($r$)$ is regular because is

generated by the regular expression r$=($a$\backslash$cup b$)$bb$.$ Really, L$($r$)=($L$($a$\backslash$cup b$))$L$($b$)$L$($b$)]=$x$\{$b$\}\{$b$\},$ x is any

word $($string$)$ in a and b$.$