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Three balls are drawn in succession without replacement from a box containing 6 red balls and 4 black balls.

a. Show all possible sample points in the sample space.
b. Define a random variable X which denotes the number of black balls
c. Determine the probability distribution of X

1 answer

Let's analyze the problem step by step.

Part a: Sample Space

In the box, there are 6 red balls (R) and 4 black balls (B). We will draw 3 balls in succession without replacement. To find all possible sample points in the sample space, we need to consider all possible combinations of the drawn balls.

Each ball can either be red (R) or black (B). When drawing 3 balls, we can represent the outcome as a sequence of the colors of the drawn balls, where each of the three positions in the sequence can be filled with either "R" or "B".

The sample space will be all the different combinations of 3 draws from the total of 10 balls. The possible outcomes are:

  1. RRR
  2. RRB
  3. RBR
  4. BRR
  5. RBB
  6. BRB
  7. BBR
  8. BBB
  9. BBR
  10. RRBR
  11. RBBR
  12. RBRB
  13. RRR
  14. RRB
  15. RBB
  16. BB

However, there are duplicates in the counting of each arrangement based on different sequences, so let's list each case distinctly. The distinct combinations of arrangements based on the number of black balls drawn (which can be 0, 1, 2, or 3) are:

  1. 0 Black Balls (3 Red Balls): RRR
  2. 1 Black Ball (2 Red Balls): RRB, RBR, BRR
  3. 2 Black Balls (1 Red Ball): RBB, BRB, BBR
  4. 3 Black Balls: BBB

Thus, the full sample space is:

  • RRR
  • RRB
  • RBR
  • BRR
  • RBB
  • BRB
  • BBR
  • BBB

Part b: Random Variable X

Let \( X \) be defined as the number of black balls drawn in the 3 draws. Therefore, \( X \) can take the following values:

  • \( X = 0 \): No black balls are drawn.
  • \( X = 1 \): One black ball is drawn.
  • \( X = 2 \): Two black balls are drawn.
  • \( X = 3 \): Three black balls are drawn.

Part c: Probability Distribution of X

To calculate the probabilities, we will determine how many outcomes correspond to each value of \( X \):

  1. For \( X = 0 \) (No black balls drawn):

    • Only outcome: RRR
    • Number of outcomes: 1
    • Probability: \( P(X = 0) = \frac{1}{\binom{10}{3}} = \frac{1}{120} \)

  2. For \( X = 1 \) (One black ball drawn):

    • Outcomes: RRB, RBR, BRR
    • Number of outcomes: 3
    • Probability: \( P(X = 1) = \frac{3}{\binom{10}{3}} = \frac{3}{120} = \frac{1}{40} \)

  3. For \( X = 2 \) (Two black balls drawn):

    • Outcomes: RBB, BRB, BBR
    • Number of outcomes: 3
    • Probability: \( P(X = 2) = \frac{3}{\binom{10}{3}} = \frac{3}{120} = \frac{1}{40} \)

  4. For \( X = 3 \) (Three black balls drawn):

    • Only outcome: BBB
    • Number of outcomes: 1
    • Probability: \( P(X = 3) = \frac{1}{\binom{10}{3}} = \frac{1}{120} \)

Summary of the Probability Distribution of \( X \):

The probability distribution of \( X \) can be summarized as follows:

  • \( P(X = 0) = \frac{1}{10} \)
  • \( P(X = 1) = \frac{3}{10} \)
  • \( P(X = 2) = \frac{3}{10} \)
  • \( P(X = 3) = \frac{1}{10} \)

Conclusion

Thus, the probabilities associated with each outcome of the random variable \( X \) are as follows:

  • \( P(X = 0) = \frac{1}{10} \)
  • \( P(X = 1) = \frac{3}{10} \)
  • \( P(X = 2) = \frac{3}{10} \)
  • \( P(X = 3) = \frac{1}{10} \)