Probability

One bag contains $5$ white and $3$ black marbles. Another bag contains $4$ white and $6$ black marbles. If one marble is drawn from each bag, find the probability that

1. both are white;
2. both are black;
3. one is white and one is black.

Let bag $B_1$ contain $5$ white and $3$ black marbles.

Let bag $B_2$ contain $4$ white and $6$ black marbles.

1. Let $A$ be the event of drawing a white marble from bag $B_1$
   
   Let $B$ be the event of drawing a white marble from bag $B_2$
   
   Then, probability of drawing white marbles from both the bags $= P \left(A \cap B\right)$
   
   $\because \;$ $A$ and $B$ are independent events, $P \left(A \cap B\right) = P \left(A\right) \times P \left(B\right)$
   
   Now, $P \left(A\right) = \dfrac{5}{8}$, $\;$ $P \left(B\right) = \dfrac{4}{10}$
   
   $\therefore \;$ $P \left(A \cap B\right) = \dfrac{5}{8} \times \dfrac{4}{10} = \dfrac{1}{4}$
   
   
2. Let $C$ be the event of drawing a black marble from bag $B_1$
   
   Let $D$ be the event of drawing a black marble from bag $B_2$
   
   Then, probability of drawing black marbles from both the bags $= P \left(C \cap D\right)$
   
   $\because \;$ $C$ and $D$ are independent events, $P \left(C \cap D\right) = P \left(C\right) \times P \left(D\right)$
   
   Now, $P \left(C\right) = \dfrac{3}{8}$, $\;$ $P \left(D\right) = \dfrac{6}{10}$
   
   $\therefore \;$ $P \left(C \cap D\right) = \dfrac{3}{8} \times \dfrac{6}{10} = \dfrac{9}{40}$
   
   
3. Let $E$ be the event of drawing a white marble from bag $B_1$
   
   Let $F$ be the event of drawing a black marble from bag $B_2$
   
   Let $G$ be the event of drawing a black marble from bag $B_1$
   
   Let $H$ be the event of drawing a white marble from bag $B_2$
   
   Then, probability of drawing a white marble from one bag and a black marble from the other bag $= P \left(E \cap F\right) + P \left(G \cap H\right)$
   
   $\because \;$ $E$, $F$ are independent events and $G$, $H$ are independent events, we have
   
   $= P \left(E \cap F\right) + P \left(G \cap H\right) = P \left(E\right) \times P \left(F\right) + P \left(G\right) \times P \left(H\right)$
   
   Now, $P \left(E\right) = \dfrac{5}{8}$, $\;$ $P \left(F\right) = \dfrac{6}{10}$, $\;$ $P \left(G\right) = \dfrac{3}{8}$, $\;$ $P \left(F\right) = \dfrac{4}{10}$
   
   $\begin{aligned} \therefore \; P \left(E \cap F\right) + P \left(G \cap H\right) & = \dfrac{5}{8} \times \dfrac{6}{10} + \dfrac{3}{8} \times \dfrac{4}{10} \\\\ & = \dfrac{42}{80} \\\\ & = \dfrac{21}{40} \end{aligned}$