Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Let's solve this problem step by step.
We have a box containing four red balls and eight black balls, making a total of twelve balls.
### Finding [tex]\( P(B) \)[/tex]
1. Probability of choosing a black ball first ([tex]\( P(B) \)[/tex]):
Total number of balls = 12
Number of black balls = 8
The probability of choosing a black ball first is given by the ratio of the number of black balls to the total number of balls.
[tex]\[ P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{8}{12} = \frac{2}{3} \approx 0.6667 \][/tex]
### Finding [tex]\( P(R \mid B) \)[/tex]
2. Probability of choosing a red ball second given that the first ball was black ([tex]\( P(R \mid B) \)[/tex]):
After choosing a black ball, one black ball is removed from the total, leaving us with 11 balls in total (4 red balls and 7 black balls remaining).
The probability of choosing a red ball out of these remaining balls is given by the ratio of the number of red balls to the new total number of balls.
[tex]\[ P(R \mid B) = \frac{\text{Number of red balls}}{\text{Remaining number of balls}} = \frac{4}{11} \approx 0.3636 \][/tex]
### Finding [tex]\( P(B \cap R) \)[/tex]
3. Probability of both events happening ([tex]\( P(B \cap R) \)[/tex]):
The probability of both events happening is the product of the probability of each individual event happening sequentially. This is the joint probability of choosing a black ball first and a red ball second.
[tex]\[ P(B \cap R) = P(B) \times P(R \mid B) \][/tex]
Substituting in the values we found earlier:
[tex]\[ P(B \cap R) = \left(\frac{2}{3}\right) \times \left(\frac{4}{11}\right) \approx 0.6667 \times 0.3636 \approx 0.2424 \][/tex]
### Converting to Percentage
4. Convert the probability to a percentage:
[tex]\[ P(B \cap R) \approx 0.2424 \][/tex]
To convert this to a percentage, multiply by 100:
[tex]\[ P(B \cap R) \times 100 \approx 24.24\% \][/tex]
### Summary
- The probability of choosing a black ball first ([tex]\( P(B) \)[/tex]) is approximately 0.6667.
- The probability of choosing a red ball second given the first was black ([tex]\( P(R \mid B) \)[/tex]) is approximately 0.3636.
- The probability of both events happening ([tex]\( P(B \cap R) \)[/tex]) is approximately 0.2424.
- The probability that the first ball chosen is black and the second ball chosen is red is about 24.24 percent.
We have a box containing four red balls and eight black balls, making a total of twelve balls.
### Finding [tex]\( P(B) \)[/tex]
1. Probability of choosing a black ball first ([tex]\( P(B) \)[/tex]):
Total number of balls = 12
Number of black balls = 8
The probability of choosing a black ball first is given by the ratio of the number of black balls to the total number of balls.
[tex]\[ P(B) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{8}{12} = \frac{2}{3} \approx 0.6667 \][/tex]
### Finding [tex]\( P(R \mid B) \)[/tex]
2. Probability of choosing a red ball second given that the first ball was black ([tex]\( P(R \mid B) \)[/tex]):
After choosing a black ball, one black ball is removed from the total, leaving us with 11 balls in total (4 red balls and 7 black balls remaining).
The probability of choosing a red ball out of these remaining balls is given by the ratio of the number of red balls to the new total number of balls.
[tex]\[ P(R \mid B) = \frac{\text{Number of red balls}}{\text{Remaining number of balls}} = \frac{4}{11} \approx 0.3636 \][/tex]
### Finding [tex]\( P(B \cap R) \)[/tex]
3. Probability of both events happening ([tex]\( P(B \cap R) \)[/tex]):
The probability of both events happening is the product of the probability of each individual event happening sequentially. This is the joint probability of choosing a black ball first and a red ball second.
[tex]\[ P(B \cap R) = P(B) \times P(R \mid B) \][/tex]
Substituting in the values we found earlier:
[tex]\[ P(B \cap R) = \left(\frac{2}{3}\right) \times \left(\frac{4}{11}\right) \approx 0.6667 \times 0.3636 \approx 0.2424 \][/tex]
### Converting to Percentage
4. Convert the probability to a percentage:
[tex]\[ P(B \cap R) \approx 0.2424 \][/tex]
To convert this to a percentage, multiply by 100:
[tex]\[ P(B \cap R) \times 100 \approx 24.24\% \][/tex]
### Summary
- The probability of choosing a black ball first ([tex]\( P(B) \)[/tex]) is approximately 0.6667.
- The probability of choosing a red ball second given the first was black ([tex]\( P(R \mid B) \)[/tex]) is approximately 0.3636.
- The probability of both events happening ([tex]\( P(B \cap R) \)[/tex]) is approximately 0.2424.
- The probability that the first ball chosen is black and the second ball chosen is red is about 24.24 percent.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
