The probability for choosing green and red is
![(1)/(12)](https://img.qammunity.org/2020/formulas/mathematics/college/apapam18poccy9t7a80qnl67y6ew26hwtb.png)
Solution:
Given that , A bag contains 1 red, 1 yellow, 1 blue, and 1 green marble.
We have to find what is the probability of choosing a green marble, not replacing it, and then choosing a red marble?
Now, we know that,
![\text { probability of an event }=\frac{\text { number of favourable outcomes }}{\text { total number of outcomes }}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/tmjr4wuc5p55i0xce4h5bvicses99cammp.png)
So, total possible outcomes = 1 red + 1 yellow + 1 blue + 1 green = 4
![\text { Now, probability for green marble }=\frac{1 \text { green marble }}{4 \text { marbles }}=(1)/(4)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/d9297cbpptr8ytcxotmtj7v41ywsfk5h08.png)
And now, total outcomes will be only 3 as we are not replacing the picked marble.
![\begin{array}{l}{\text { Then, probability for red marble }=\frac{1 \text { red marble }}{3 \text { marbles }}=(1)/(3)} \\\\ {\text { Then overall probability }=(1)/(4) * (1)/(3)=(1)/(12)}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/m3e0vaxmx4ptc0zu02e887m4n61bdeerc9.png)
Hence, the probability for choosing green and red is
![(1)/(12)](https://img.qammunity.org/2020/formulas/mathematics/college/apapam18poccy9t7a80qnl67y6ew26hwtb.png)