1. So the question is P(first even then not 2)
2. P(first even then not 2)= [tex] \frac{n(first even then not 2)}{n(sample space)} [/tex]
3. n(first even then not 2)=n{(2,1),(2,3),(2,4),(2,5),(2,6),(4,1),(4,3),(4,4),(4,5),(4,6),(6,1),(6,3),(6,4),(6,5),(6,6)}=15
4. n(sample space)= 6(possible outcomes for the first throw)*6(for the second throw)=36
5. P(first even then not 2)=[tex] \frac{15}{36} [/tex]
