To find $P(B)$, the probability that the two balls drawn are of different colors, we proceed by calculating the probability directly.

Step-by-Step Solution

1. Identify the total number of ways to draw two balls from the urn.

   • The urn has a total of $6 + 4 = 10$ balls.
   • The number of ways to choose 2 balls from 10 is: $\binom{10}{2} = \frac{10 \times 9}{2} = 45$

2. Determine the outcomes where the two balls are of different colors (event $B$).

   • There are two cases to consider:
     1. The first ball is red and the second ball is blue.
     2. The first ball is blue and the second ball is red.

3. Calculate the probability of drawing one red and one blue ball.

   • The number of ways to select one red ball and one blue ball from 6 red and 4 blue is: $6 \times 4 = 24$
   • Thus, there are 24 outcomes in which the two balls drawn are of different colors.

4. Calculate $P(B)$.

   • $P(B)$ is the ratio of the favorable outcomes for event $B$ (different colors) to the total possible outcomes: $P(B) = \frac{24}{45} = \frac{8}{15}$

Answer

$P(B) = \frac{8}{15}$

Would you like a further breakdown or explanation of this solution?

Here are some related questions to consider:

1. What is $P(A)$, the probability of drawing at least one red ball?
2. What is $P(A \cap B)$, the probability that both events $A$ and $B$ occur?
3. What is $P(\text{both balls are the same color})$ in this scenario?
4. How would the probability $P(B)$ change if the draws were with replacement?
5. How does the calculation change if the number of balls in the urn is increased?

Tip: When dealing with probabilities in "without replacement" scenarios, always remember that the number of total outcomes decreases after each draw.