Denote by [tex]B_1,B_2[/tex] the event that a marble is drawn from box 1 or 2, respectively, and by [tex]R[/tex] the event that a red marble is drawn. If a blue marble is drawn, we'll denote that by the complement of [tex]R[/tex], or [tex]R^C[/tex].
a. If a red marble is drawn, either it's drawn from box 1 or box 2. The events [tex]B_1[/tex] and [tex]B_2[/tex] are mutually exclusive, so we have
[tex]\mathbb P(R)=\mathbb P((R\cap B_1)\cup(R\cap B_2))=\mathbb P(R\cap B_1)+\mathbb P(R\cap B_2)[/tex]
Now invoke the definition of conditional probability:
[tex]\mathbb P(R)=\mathbb P(B_1)\mathbb P(R\mid B_1)+\mathbb P(B_2)\mathbb P(R\mid B_2)[/tex]
We're explicitly given the probability of drawing from either box. We also know the probabilities of drawing a red marble from box 1 or 2 are, respectively,
[tex]\mathbb P(R\mid B_1)=\dfrac{\dbinom41\dbinom80}{\dbinom{12}1}=\dfrac4{12}=\dfrac13[/tex]
and
[tex]\mathbb P(R\mid B_2)=\dfrac{\dbinom51\dbinom30}{\dbinom81}=\dfrac58[/tex]
So
[tex]\mathbb P(R)=\dfrac25\times\dfrac13+\dfrac35\times\dfrac58=\dfrac{61}{120}\approx0.508[/tex]
b. Given that we draw a red marble, we're looking for the probability that it was drawn from box 1, i.e. [tex]\mathbb P(B_1\mid R)[/tex]. We use Bayes' theorem:
[tex]\mathbb P(B_1\mid R)=\dfrac{\mathbb P(B_1\cap R)}{\mathbb P(R)}=\dfrac{\mathbb P(B_1)\mathbb P(R\mid B_1)}{\mathbb P(R)}[/tex]
which is really just a matter of using the definition of conditional probability to rewrite the numerator on the the right side, as
[tex]\mathbb P(R\cap B_1)=\mathbb P(R)\mathbb P(B_1\mid R)=\mathbb P(B_1)\mathbb P(R\mid B_1)[/tex]
We know [tex]\mathbb P(R)[/tex] from part (a), so we compute
[tex]\mathbb P(B_1\mid R)=\dfrac{\dfrac25\times\dfrac13}{\dfrac{61}{120}}=\dfrac{16}{61}\approx0.262[/tex]
c. Now we look for [tex]\mathbb P(B_1\mid R^C)[/tex], which can be computed similarly.
[tex]\mathbb P(B_1\mid R^C)=\dfrac{\mathbb P(B_1\cap R^C)}{\mathbb P(R^C)}=\dfrac{\mathbb P(B_1)\mathbb P(R^C\mid B_1)}{1-\mathbb P(R)}[/tex]
[tex]\mathbb P(B_1\mid R^C)=\dfrac{\dfrac25\times\dfrac23}{1-\dfrac{61}{120}}=\dfrac{32}{59}\approx0.542[/tex]
