Step-by-step explanation:
The total number of coin in the first bag is given below as
![\begin{gathered} 6+2+7+10=25 \\ n(S)=25 \\ n(R)=6 \\ N(Si)=2 \\ n(O)=10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/iyws1yvmyvcnm4gf7rt0vmlr9fn9xsvu3o.png)
The probabaility of picking an orange ball from the first ba will b calcuated below as
![\begin{gathered} Pr(O)=(n(O))/(n(S)) \\ Pr(O)=(10)/(25)=(2)/(5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6l0n6s3s5ovta8r188eebez864bqnb1gyv.png)
Step 2:
The total number of coin in the second bag is given below as
![\begin{gathered} 4+12+5+3=24 \\ n(S)=24 \\ n(R)=4 \\ n(Si)=12 \\ n(Y)=5 \\ n(O)=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gcutltemszlrp43aqq2dzlfzizbg07gd1s.png)
The probabaility of picking an orange ball from the second bag will be calcuated below as
![\begin{gathered} Pr(O)=(n(O))/(n(S)) \\ Pr(O)=(3)/(24)=(1)/(8) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jxo5dyry42p7yqkh19zoqmwu1cwp27eioi.png)
Hence,
The probabaility of picking an orange coin from both bags will be calculated below as
![\begin{gathered} Pr(O_1,O_2)=(2)/(5)*(1)/(8)=(2)/(40)=0.05 \\ Pr(O)=0.05*100=5\% \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/48w3ocspqfi3ha0pf4ylbs9maz6syrg4i9.png)
Hence,
The final answer is
![5\%](https://img.qammunity.org/2023/formulas/mathematics/college/b0t441zzg97jqnhwgc2nqnvwa52mye3slq.png)