Solution:
Let white balls be w, and black balls be b;
In the first bag;
![n(w)=5,n(b)=3,n(T)=8](https://img.qammunity.org/2023/formulas/mathematics/college/zcmpkayily12wehxdozycz5sm2yrxno8e2.png)
But, in the second bag;
![n(w)=4,n(b)=6,n(T)=10](https://img.qammunity.org/2023/formulas/mathematics/college/m50qahbgiktich2acfrwv7x67jylv0vye6.png)
(i) The probability that both balls drawn from each bag are white is;
![\begin{gathered} P(w\text{ and }w)=(5)/(8)*(4)/(10) \\ P(w\text{ and }w)=(20)/(80) \\ P(w\text{ and }w)=(1)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/c6gsodhtn5rvlnlhc5sj084fvdvhwllifo.png)
(ii) The probability that both balls drawn from each bag are black is;
![\begin{gathered} P(b\text{ and }b)=(3)/(8)*(6)/(10) \\ P(b\text{ and }b)=(18)/(80) \\ P(b\text{ and }b)=(9)/(40) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4zq55at7lwz1x8kfbry663oa4pfgb0e65q.png)
(iii) The probability that one is white and another is black is;
![\begin{gathered} P(w\text{ and }b)=((5)/(8)*(6)/(10))+((3)/(8)*(4)/(10)) \\ P(w\text{ and }b)=(30)/(80)+(12)/(80) \\ P(w\text{ and }b)=(42)/(80) \\ P(w\text{ and }b)=(21)/(40) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jll9eugyrhv2979rd0ru8tsa5v7lgykpb2.png)