Solution:
Given:
[tex]\begin{gathered} Red,R=8 \\ Yellow,Y=4 \\ Total=12 \end{gathered}[/tex]
The probability that 1 marble will be yellow and 1 marble will be red is;
[tex]\begin{gathered} P(YR)\text{ or }P(RY)=(\frac{4}{12}\times\frac{8}{11})+(\frac{8}{12}\times\frac{4}{11}) \\ P(YR)\text{ or }P(RY)=\frac{32}{132}+\frac{32}{132} \\ P(YR)\text{ or }P(RY)=\frac{64}{132} \\ P(YR)\text{ or }P(RY)=\frac{16}{33} \end{gathered}[/tex]
Therefore, the probability that 1 marble will be yellow and 1 marble will be red is
