At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our 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]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
