Solution:
The probability of an event is expressed as
![P(event)=\frac{number\text{ of desirable outcome}}{number\text{ of possible outcomes}}](https://img.qammunity.org/2023/formulas/mathematics/college/zes67bknvvjgpmtr6kv9gzh9p2amgf512d.png)
Given that:
![\begin{gathered} blue\text{ marbes:6} \\ red\text{ marbles:7} \\ orange\text{ marbles:4} \\ green\text{ marbles:3} \\ total\text{ number of marbles: 20} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qf2wk5sb7l517bb56o256xym3iabdq87es.png)
This implies that the number of possible outcomes is 20.
Provided that once a marble is drawn, it's replaced, the probability of selecting a red marble then a blue marble is expressed as
![P(R\text{ and B\rparen=P\lparen red for first selection\rparen}*\text{P\lparen blue for second selection\rparen}](https://img.qammunity.org/2023/formulas/mathematics/college/ym8siziuay7n75jdfob8szvbvjdypelsqd.png)
where
![\begin{gathered} P(red\text{ for first selection\rparen=}\frac{number\text{ of red balls}}{total\text{ number of balls}}=(7)/(20) \\ \Rightarrow P(red\text{ for first selection\rparen=}(7)/(20) \\ P(blue\text{ for second selection\rparen=}\frac{number\text{ of blue balls}}{total\text{ number of balls left}}=(6)/(20-1) \\ \Rightarrow P(blue\text{ for second selection\rparen=}(6)/(19) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w420myu86twop5n1kmwvduvzfn6ad14g28.png)
Thus, we have
![\begin{gathered} P(Red\text{ and Blue\rparen=}(7)/(20)*(6)/(19) \\ =(21)/(190) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8r082aw6xokrstkhs7yblec2eqz3xlx4s7.png)
Hence, the probability of selecting a red then a blue marble is
![(21)/(190)](https://img.qammunity.org/2023/formulas/mathematics/college/xdoqo85h86x0svrzojwgajp9vk6ssi6fus.png)