Total spins = 11 + 10 +9 = 30
Total spins of red = 11
Total spins of green = 10
Total spins of blue = 9
The formula for finding probability, P, is
![P=\frac{\text{required outcomes}}{possible\text{ outcomes}}](https://img.qammunity.org/2023/formulas/mathematics/college/6lm31n2t4ak3s99lm14upuno8a3lu58e1i.png)
Let the probability of spinning a red be P(R)
Let the probability of spinning a green be P(G)
Let the probability of spinning a blue be P(B)
a) The experimental probability of spinning a green, P(R) is
![\begin{gathered} P=\frac{\text{required outcome}}{possible\text{ outcome}} \\ P(R)=\frac{\text{Total spins of green}}{Total\text{ spins}}=(10)/(30)=(1)/(3) \\ P(R)=(1)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5sht7ded0bgwyzrxf1uxw2k4tu20reuwdk.png)
Hence, the experimental probability of spinning a green is 1/3
b) The experimental probability of spinning a blue, P(B), is
![\begin{gathered} P=\frac{\text{required outcome}}{possible\text{ outcome}} \\ P(B)=\frac{\text{Total spins of green}}{Total\text{ spins}}=(9)/(30)=(3)/(10) \\ P(B)=(3)/(10) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3p7vy1a70rqsk8eg4ew5z3aqmfz91c7hye.png)
The experimental probability of spinning a blue, P(B), is 3/10
The experimental probability of spinning a green, P(G), is 1/3
The experimental probability of spinning a blue or green, P(B OR G), will be
![\begin{gathered} P(B\text{ OR G)}=P(B)+P(G) \\ P(B\text{ OR G)}=(3)/(10)+(1)/(3)=(9+10)/(30)=(19)/(30) \\ P(B\text{ OR G)}=(19)/(30) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hyq2p894spytl9ol5gs3vmdut3im83uw1h.png)
Hence, the experimental probability of spinning a blue or green, P(B OR G) is 19/30
c) The sample space for this experiment is the total possible outcome that can be obtained
Hence, the sample space for this experiment is 30