At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
