ii.) Probablity of $2$ bags containing white and black balls. 
A bag contains $4$ white balls and $2$ black balls ,
   another contains $3$ white balls and $5$ black 
  balls . If one ball is drawn from each bag, 
  determine the probability that one is white and one is black.

$\color{green}{ a.)\ \dfrac{13}{24} }\\
 b.)\ \dfrac{15}{24} \\
 c.)\ \dfrac{11}{21}  \\
 d.)\ \dfrac{1}{2} $
I did $\dfrac12 \times \dfrac46 \times \dfrac58 + \dfrac12 \times  \dfrac58 \times \dfrac46 =\dfrac{5}{12} $
But the answer is given as option $a.)$ 
I look for a short and simple way.
I have studied maths upto $12$th grade.
 A: Hint.  What is the probability that you draw a white ball from the first bag and a black ball from the second bag?  What is the probability that you draw a black ball from the first bag and a white ball from the second bag?  Just add these together.
Second Hint. There should be no $1/2$ anywhere, and no numbers $\geq 1$.
A: Let's call the bags "Bag 1" and "Bag 2", in the order you listed them. What is the probability that we draw a white ball from Bag 1 and a black ball from Bag 2? Bag 1 has $4$ white balls out of $6$ in total, so the probability of drawing a white ball is $\frac{4}{6} = \frac{2}{3}$. By the same argument, the probability of drawing a black ball from Bag 2 is $\frac{5}{8}$. By multiplying them together we get the total probability of drawing a white ball and black ball in this way: $\frac{2}{3} \cdot \frac{5}{8} = \frac{10}{24}$.
However, we must also consider the case when we draw a black ball from Bag 1 and a white ball from Bag 2. By the exact same argument as the previous case, only with slightly different numbers, you get that the probability in this case is $\frac{3}{24}$, and by adding our two cases we get the correct answer: $$\frac{10}{24} + \frac{3}{24} = \frac{13}{24}$$
A: In total you have $48$ combinations if we don't care of which color we take.
Now in those $48$ combinations we have included the combinations where we have every of the $4$ white ball of the first urn arranged with the $5$ black of the second urn.Hence we have $4\cdot 5=20$ combinations in total. 
The other  case we care about  is the combinations of the $2$ black balls of the first urn with the $3$ white balls of the second urn,so we have $6$ combinations.
Finally if we count all the combinations we have mentioned above ,we have $$\frac {20+6}{48}=\cfrac{13}{24}$$ which is the probability of drawing a white and black ball together.
A: 
A bag contains 4 white balls and 2 black balls , another contains 3 white balls and 5 black balls .

Let $A$ be the count of white balls drawn from the first bag, and $B$ the count of white balls drawn from the second.   ( These will independently be either $0$ or $1$ since we are drawing only one ball from each bag. )
So $\mathsf P(A{=}0) = \tfrac 2 6 \\ \mathsf P(A{=}1)=\tfrac 4 6 \\ \mathsf P(B{=}0)=\tfrac 5 8 \\ \mathsf P(B{=}1) =\tfrac 3 8$

If one ball is drawn from each bag, determine the probability that one is white and one is black.

We are after the probability of the event of: $A{+}B{=}1$.   Which is either $A{=}1 , B{=}0$ xor $A{=}0, B{=}1$.
$$\begin{align}\mathsf P(A{+}B{=}1) & = \mathsf P(A{=}1, B{=}0)+\mathsf P(A{=}0, B{=}1) & \because \text{ exclusive events} \\ & = \mathsf P(A{=}1)\cdot\mathsf P(B{=}0)+\mathsf P(A{=}0)\cdot\mathsf P(B{=}1) & \because \text{ independence} \\ & = \tfrac 4 6 \cdot \tfrac 5 8+ \tfrac 2 6\cdot\tfrac 3 8\end{align}$$

Note: There's no factor of $\tfrac 1 2$ because its given that its certain that we will draw one ball from each bag, and the order of draw does not matter.   The only randomness is in the colour of the ball drawn from each bag.
