Please solve Q. No. 1

Dear Student, your question seems to be incomplete. Assuming your question to be " Bag (A)  has 4 red and 5 black balls. Second bag (B) has 3 red and 7 black balls. One ball is drawn bag A and two from bag B. Find the probability that out of 3 balls drawn,two are black and one is red?"

Solution)
$$\begin{gathered} 2 \mathrm{black} \mathrm{and} \mathrm{one} \mathrm{red} \mathrm{ball} \mathrm{can} \mathrm{be} \mathrm{drawn} \mathrm{in} \mathrm{two} \mathrm{mutually} \mathrm{exclusive} \mathrm{ways}. \\ 1) \mathrm{Drawing} \mathrm{one} \mathrm{black} \mathrm{ball} \mathrm{from} \mathrm{bagA} \mathrm{and} \mathrm{two} \mathrm{balls} \mathrm{from} \mathrm{bag} \mathrm{B} \mathrm{out} \mathrm{of} \mathrm{which} \mathrm{one} \mathrm{is} \\ \mathrm{black} \mathrm{and} \mathrm{other} \mathrm{is} \mathrm{red}. \\ 2) \mathrm{Drawing} \mathrm{one} \mathrm{red} \mathrm{ball} \mathrm{from} \mathrm{bag} \mathrm{A} \mathrm{two} \mathrm{black} \mathrm{balls} \mathrm{from} \mathrm{bag} \mathrm{B}. \\ \mathrm{Thus}, \mathrm{if} \mathrm{we} \mathrm{defone} \mathrm{the} \mathrm{following} \mathrm{events}: \\ {\mathrm{E}}_1=\mathrm{Drawing} \mathrm{a} \mathrm{black} \mathrm{ball} \mathrm{from} \mathrm{bag} \mathrm{A} \\ {\mathrm{E}}_2=\mathrm{Drawing} \mathrm{one} \mathrm{red} \mathrm{and} \mathrm{one} \mathrm{black} \mathrm{ball} \mathrm{from} \mathrm{bag} \mathrm{B} \\ {\mathrm{E}}_3=\mathrm{Drawing} \mathrm{one} \mathrm{red} \mathrm{ball} \mathrm{from} \mathrm{bag} \mathrm{A} \\ {\mathrm{E}}_4=\mathrm{Drawing} \mathrm{two} \mathrm{black} \mathrm{balss} \mathrm{from} \mathrm{bag} \mathrm{B}. \\ \mathrm{Then} \mathrm{P}\left({\mathrm{E}}_1\right)=\frac{5}{9},\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{{}^3{\mathrm{C}}_1{\times }^7{\mathrm{C}}_1}{{}^{10}{\mathrm{C}}_2}=\frac{21}{45},\mathrm{P}\left({\mathrm{E}}_3\right)=\frac{4}{9},\mathrm{P}\left({\mathrm{E}}_4\right)=\frac{{}^7{\mathrm{C}}_2}{{}^{10}{\mathrm{C}}_2}=\frac{7}{15} \\ \mathrm{Now}, \\ \mathrm{P}(\mathrm{Two} \mathrm{black} \mathrm{balls} \mathrm{and} \mathrm{one} \mathrm{red} \mathrm{ball}) \\ =\mathrm{P}\left[\left(1 \mathrm{black} \mathrm{from} \mathrm{bag} \mathrm{A} \mathrm{and} \mathrm{one} \mathrm{red} \mathrm{and} \mathrm{one} \mathrm{black} \mathrm{ball} \mathrm{from} \mathrm{bag} \mathrm{B}\right)\mathrm{or} \left(\mathrm{one} \mathrm{red} \mathrm{from} \mathrm{bag} \mathrm{A} \mathrm{and} 2 \mathrm{black} \mathrm{from} \mathrm{bag} \mathrm{B}\right)\right] \\ =\mathrm{P}(\mathrm{I} \mathrm{or} \mathrm{II}) \\ =\mathrm{P}(\mathrm{I}\cup \mathrm{II}) \\ =\mathrm{P}\left[\left({\mathrm{E}}_1\cap {\mathrm{E}}_2\right)\cup \left({\mathrm{E}}_3\cap {\mathrm{E}}_4\right)\right] \\ =\mathrm{P}\left({\mathrm{E}}_1\cap {\mathrm{E}}_2\right)+\mathrm{P}\left({\mathrm{E}}_3\cap {\mathrm{E}}_4\right) \\ =\mathrm{P}\left({\mathrm{E}}_1\right)\mathrm{P}\left({\mathrm{E}}_2\right)+\mathrm{P}\left({\mathrm{E}}_3\right)\mathrm{P}\left({\mathrm{E}}_4\right) \\ =\frac{5}{9}\times \frac{21}{45}+\frac{4}{9}\times \frac{7}{15}=\frac{63}{135}=\frac{7}{15} \end{gathered}$$
Regards!