Pls give answer for 9c thanks

Answer:
probability for 3 on a dice (P3) in one dice =
[tex] \frac{1}{6} [/tex]
probability for even no. of a dice (Pe) =
[tex] \frac{3}{6} = \frac{1}{2} [/tex]
Therefore, the probability of finding 3 on one dice and even on another (P) = P3 + Pe
[tex] = \frac{1}{6} + \frac{1}{2} = \frac{1 + 3}{6} = \frac{4}{6} = \frac{2}{3} [/tex]