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Sagot :
To solve the problem of finding the probability that two balls drawn from a bag are of the same color, we need to consider the probabilities of each possible color pair being drawn. The bag contains 3 white balls, 4 green balls, and 5 red balls, making a total of 12 balls. Let’s break down the steps:
### Step-by-Step Solution
#### Total number of balls:
There are [tex]\( 3 \)[/tex] white balls, [tex]\( 4 \)[/tex] green balls, and [tex]\( 5 \)[/tex] red balls. So, the total number of balls is:
[tex]\[ 3 + 4 + 5 = 12 \][/tex]
#### Probability of drawing two white balls:
1. The probability of drawing one white ball first is:
[tex]\[ \frac{3}{12} \][/tex]
2. After drawing one white ball, there are [tex]\( 2 \)[/tex] white balls left out of [tex]\( 11 \)[/tex] total balls. Therefore, the probability of drawing the second white ball is:
[tex]\[ \frac{2}{11} \][/tex]
3. Combining these probabilities, the likelihood of drawing two white balls is:
[tex]\[ \left( \frac{3}{12} \right) \times \left( \frac{2}{11} \right) = \frac{6}{132} = \frac{1}{22} \approx 0.0455 \][/tex]
#### Probability of drawing two green balls:
1. The probability of drawing one green ball first is:
[tex]\[ \frac{4}{12} \][/tex]
2. After drawing one green ball, there are [tex]\( 3 \)[/tex] green balls left out of [tex]\( 11 \)[/tex] total balls. Therefore, the probability of drawing the second green ball is:
[tex]\[ \frac{3}{11} \][/tex]
3. Combining these probabilities, the likelihood of drawing two green balls is:
[tex]\[ \left( \frac{4}{12} \right) \times \left( \frac{3}{11} \right) = \frac{12}{132} = \frac{1}{11} \approx 0.0909 \][/tex]
#### Probability of drawing two red balls:
1. The probability of drawing one red ball first is:
[tex]\[ \frac{5}{12} \][/tex]
2. After drawing one red ball, there are [tex]\( 4 \)[/tex] red balls left out of [tex]\( 11 \)[/tex] total balls. Therefore, the probability of drawing the second red ball is:
[tex]\[ \frac{4}{11} \][/tex]
3. Combining these probabilities, the likelihood of drawing two red balls is:
[tex]\[ \left( \frac{5}{12} \right) \times \left( \frac{4}{11} \right) = \frac{20}{132} = \frac{5}{33} \approx 0.1515 \][/tex]
#### Total probability of drawing two balls of the same color:
To find the total probability, we add the probabilities of each color pair:
[tex]\[ 0.0455 + 0.0909 + 0.1515 = 0.288 \][/tex]
#### Converting the probability to a percentage:
To express this probability as a percentage, we multiply by 100:
[tex]\[ 0.288 \times 100 = 28.8\% \][/tex]
Thus, the probability that the two balls drawn are of the same color is approximately [tex]\( 28.8\% \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{29\%} \][/tex]
So, the most appropriate answer choice is:
[tex]\[ \text{D. } 29\% \][/tex]
### Step-by-Step Solution
#### Total number of balls:
There are [tex]\( 3 \)[/tex] white balls, [tex]\( 4 \)[/tex] green balls, and [tex]\( 5 \)[/tex] red balls. So, the total number of balls is:
[tex]\[ 3 + 4 + 5 = 12 \][/tex]
#### Probability of drawing two white balls:
1. The probability of drawing one white ball first is:
[tex]\[ \frac{3}{12} \][/tex]
2. After drawing one white ball, there are [tex]\( 2 \)[/tex] white balls left out of [tex]\( 11 \)[/tex] total balls. Therefore, the probability of drawing the second white ball is:
[tex]\[ \frac{2}{11} \][/tex]
3. Combining these probabilities, the likelihood of drawing two white balls is:
[tex]\[ \left( \frac{3}{12} \right) \times \left( \frac{2}{11} \right) = \frac{6}{132} = \frac{1}{22} \approx 0.0455 \][/tex]
#### Probability of drawing two green balls:
1. The probability of drawing one green ball first is:
[tex]\[ \frac{4}{12} \][/tex]
2. After drawing one green ball, there are [tex]\( 3 \)[/tex] green balls left out of [tex]\( 11 \)[/tex] total balls. Therefore, the probability of drawing the second green ball is:
[tex]\[ \frac{3}{11} \][/tex]
3. Combining these probabilities, the likelihood of drawing two green balls is:
[tex]\[ \left( \frac{4}{12} \right) \times \left( \frac{3}{11} \right) = \frac{12}{132} = \frac{1}{11} \approx 0.0909 \][/tex]
#### Probability of drawing two red balls:
1. The probability of drawing one red ball first is:
[tex]\[ \frac{5}{12} \][/tex]
2. After drawing one red ball, there are [tex]\( 4 \)[/tex] red balls left out of [tex]\( 11 \)[/tex] total balls. Therefore, the probability of drawing the second red ball is:
[tex]\[ \frac{4}{11} \][/tex]
3. Combining these probabilities, the likelihood of drawing two red balls is:
[tex]\[ \left( \frac{5}{12} \right) \times \left( \frac{4}{11} \right) = \frac{20}{132} = \frac{5}{33} \approx 0.1515 \][/tex]
#### Total probability of drawing two balls of the same color:
To find the total probability, we add the probabilities of each color pair:
[tex]\[ 0.0455 + 0.0909 + 0.1515 = 0.288 \][/tex]
#### Converting the probability to a percentage:
To express this probability as a percentage, we multiply by 100:
[tex]\[ 0.288 \times 100 = 28.8\% \][/tex]
Thus, the probability that the two balls drawn are of the same color is approximately [tex]\( 28.8\% \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{29\%} \][/tex]
So, the most appropriate answer choice is:
[tex]\[ \text{D. } 29\% \][/tex]
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