A Box Contains Four Red Balls And Eight Black Balls. Two Balls Are Randomly Chosen From The Box And Are Not Replaced. Let Event B Be Choosing A Black Ball First And Event R Be Choosing A Red Ball Second.What Are The Following Probabilities?$\[ P(B)

Introduction

In probability theory, we often encounter scenarios where we need to calculate the likelihood of certain events occurring. In this article, we will explore a classic problem involving a box containing four red balls and eight black balls. We will examine the probabilities associated with choosing a black ball first and then a red ball second, and vice versa.

The Problem

A box contains four red balls and eight black balls. Two balls are randomly chosen from the box and are not replaced. Let event B be choosing a black ball first and event R be choosing a red ball second. We want to find the probabilities of these events occurring.

Calculating Probabilities

To calculate the probabilities, we need to consider the total number of possible outcomes and the number of favorable outcomes for each event.

Total Number of Possible Outcomes

When two balls are chosen from the box without replacement, the total number of possible outcomes is the number of ways to choose 2 balls from 12 (4 red + 8 black). This can be calculated using the combination formula:

$$\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2} = 66$$

Probability of Choosing a Black Ball First (Event B)

The probability of choosing a black ball first is the number of favorable outcomes (choosing a black ball) divided by the total number of possible outcomes:

$$P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{8}{12} = \frac{2}{3}$$

Probability of Choosing a Red Ball Second (Event R)

After choosing a black ball first, there are 11 balls left in the box, of which 4 are red. The probability of choosing a red ball second is the number of favorable outcomes (choosing a red ball) divided by the total number of possible outcomes:

$$P(R) = \frac{\text{Number of red balls}}{\text{Total number of balls remaining}} = \frac{4}{11}$$

Probability of Choosing a Black Ball First and a Red Ball Second (Event BR)

To find the probability of choosing a black ball first and a red ball second, we multiply the probabilities of each event:

$$P(BR) = P(B) \times P(R) = \frac{2}{3} \times \frac{4}{11} = \frac{8}{33}$$

Probability of Choosing a Red Ball First and a Black Ball Second (Event RB)

To find the probability of choosing a red ball first and a black ball second, we need to consider the number of favorable outcomes for each event. The probability of choosing a red ball first is:

$$P(R) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{4}{12} = \frac{1}{3}$$

After choosing a red ball first, there are 11 balls left in the box, of which 8 are black. The probability of choosing a black ball second is:

$$P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls remaining}} = \frac{8}{11}$$

The probability of choosing a red ball first and a black ball second is:

$$P(RB) = P(R) \times P(B) = \frac{1}{3} \times \frac{8}{11} = \frac{8}{33}$$

Conclusion

In this article, we calculated the probabilities associated with choosing a black ball first and a red ball second, and vice versa, from a box containing four red balls and eight black balls. We found that the probability of choosing a black ball first and a red ball second is $\frac{8}{33}$, and the probability of choosing a red ball first and a black ball second is also $\frac{8}{33}$. These results demonstrate the importance of considering the total number of possible outcomes and the number of favorable outcomes when calculating probabilities.

References

• [1] "Probability Theory" by E.T. Jaynes
• [2] "A First Course in Probability" by Sheldon Ross

Further Reading

• [1] "Conditional Probability" by Khan Academy
• [2] "Probability and Statistics" by MIT OpenCourseWare
  A Box of Red and Black Balls: Understanding Probabilities - Q&A ===========================================================

Introduction

In our previous article, we explored a classic problem involving a box containing four red balls and eight black balls. We calculated the probabilities associated with choosing a black ball first and a red ball second, and vice versa. In this article, we will answer some frequently asked questions related to this problem.

Q&A

Q: What is the total number of possible outcomes when two balls are chosen from the box without replacement?

A: The total number of possible outcomes is the number of ways to choose 2 balls from 12 (4 red + 8 black). This can be calculated using the combination formula:

$$\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2} = 66$$

Q: What is the probability of choosing a black ball first?

A: The probability of choosing a black ball first is the number of favorable outcomes (choosing a black ball) divided by the total number of possible outcomes:

$$P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{8}{12} = \frac{2}{3}$$

Q: What is the probability of choosing a red ball second after choosing a black ball first?

A: After choosing a black ball first, there are 11 balls left in the box, of which 4 are red. The probability of choosing a red ball second is the number of favorable outcomes (choosing a red ball) divided by the total number of possible outcomes:

$$P(R) = \frac{\text{Number of red balls}}{\text{Total number of balls remaining}} = \frac{4}{11}$$

Q: What is the probability of choosing a black ball first and a red ball second?

A: To find the probability of choosing a black ball first and a red ball second, we multiply the probabilities of each event:

$$P(BR) = P(B) \times P(R) = \frac{2}{3} \times \frac{4}{11} = \frac{8}{33}$$

Q: What is the probability of choosing a red ball first and a black ball second?

A: To find the probability of choosing a red ball first and a black ball second, we need to consider the number of favorable outcomes for each event. The probability of choosing a red ball first is:

$$P(R) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{4}{12} = \frac{1}{3}$$

After choosing a red ball first, there are 11 balls left in the box, of which 8 are black. The probability of choosing a black ball second is:

$$P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls remaining}} = \frac{8}{11}$$

The probability of choosing a red ball first and a black ball second is:

$$P(RB) = P(R) \times P(B) = \frac{1}{3} \times \frac{8}{11} = \frac{8}{33}$$

Q: Why do the probabilities of choosing a black ball first and a red ball second and choosing a red ball first and a black ball second differ?

A: The probabilities differ because the order of events matters. When we choose a black ball first and a red ball second, the probability of choosing a red ball second is different from the probability of choosing a red ball first and a black ball second.

Q: Can we use the same formula to calculate the probability of choosing a red ball first and a black ball second?

A: No, we cannot use the same formula to calculate the probability of choosing a red ball first and a black ball second. The formula for the probability of choosing a red ball first is:

$$P(R) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{4}{12} = \frac{1}{3}$$

After choosing a red ball first, there are 11 balls left in the box, of which 8 are black. The probability of choosing a black ball second is:

$$P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls remaining}} = \frac{8}{11}$$

The probability of choosing a red ball first and a black ball second is:

$$P(RB) = P(R) \times P(B) = \frac{1}{3} \times \frac{8}{11} = \frac{8}{33}$$

Conclusion

In this article, we answered some frequently asked questions related to the problem of choosing a black ball first and a red ball second, and vice versa, from a box containing four red balls and eight black balls. We hope that this Q&A article has provided you with a better understanding of the problem and its solutions.

References

• [1] "Probability Theory" by E.T. Jaynes
• [2] "A First Course in Probability" by Sheldon Ross

Further Reading

• [1] "Conditional Probability" by Khan Academy
• [2] "Probability and Statistics" by MIT OpenCourseWare