To solve this problem, we need to calculate the probability of drawing two red balls in succession from a bag containing 4 red balls and 5 black balls, without replacement.

Step 1: Total number of balls

The total number of balls in the bag is: $4 \, (\text{red balls}) + 5 \, (\text{black balls}) = 9 \, \text{balls}.$

Step 2: Probability of drawing the first red ball

The probability of drawing the first red ball is the number of red balls divided by the total number of balls: $P(\text{first red}) = \frac{4}{9}.$

Step 3: Probability of drawing the second red ball

Since the ball is not replaced after the first draw, there are now 3 red balls left and 8 balls in total. So, the probability of drawing a second red ball after the first red ball is: $P(\text{second red | first red}) = \frac{3}{8}.$

Step 4: Overall probability

The overall probability of drawing two red balls in succession (without replacement) is the product of the two individual probabilities: $P(\text{two red balls}) = P(\text{first red}) \times P(\text{second red | first red}) = \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6}.$

Thus, the probability of drawing two red balls in succession without replacement is: $\boxed{\frac{1}{6}}.$

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Would you like a more detailed breakdown of the process or have any other questions?

Here are 5 related questions for you to explore:

1. What would the probability be if the ball were replaced after the first draw?
2. How does the probability change if there were 6 red balls and 3 black balls in the bag?
3. How would the calculation differ if you were drawing 3 balls instead of 2?
4. What is the concept of conditional probability, and how does it apply to this problem?
5. How would the probability change if you drew the balls with replacement instead of without?

Tip: The key difference between drawing with and without replacement is how the total number of balls changes after each draw.