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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.
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