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Sagot :
Let's go through each part of the questions along with their detailed solutions step by step.
#### 1. What is the probability that someone is from the U.S?
- From the given data table:
- People from outside the U.S are [7, 10, 11, 13, 41].
- Total people are [15, 25, 16, 19, 75].
- Calculate the total number of people from outside the U.S: [tex]\(7 + 10 + 11 + 13 + 41 = 82\)[/tex].
- Calculate the total number of people: [tex]\(15 + 25 + 16 + 19 + 75 = 150\)[/tex].
- Calculate the total number of people from the U.S: [tex]\(150 - 82 = 68\)[/tex].
- Probability [tex]\(P(U.S)\)[/tex] is:
[tex]\[ P(U.S) = \frac{\text{Total people from the U.S}}{\text{Total people}} = \frac{68}{150} = 0.4533 \][/tex]
#### 2. What is the probability that someone has at least 2 children?
- Number of people with 2 or more children is: 16 (with 2 children) + 19 (with 3 children) + 75 (with 4 or more children)
- [tex]\(P(\text{At least 2 children})\)[/tex] is:
[tex]\[ P(\text{At least 2 children}) = \frac{16 + 19 + 75}{150} = \frac{110}{150} = 0.7333 \][/tex]
#### 3. What is the probability that someone is from the U.S and has one child?
- Number of people with 1 child: 25 (Total), out of which 10 are from outside the U.S, so 25 - 10 = 15 are from the U.S.
- Probability [tex]\(P(U.S \cap 1 \text{ child})\)[/tex] is:
[tex]\[ P(U.S \cap 1 \text{ child}) = \frac{15}{150} = 0.1 \][/tex]
#### 4. What is the probability that someone is from the U.S or has 0 children?
- Number of people with 0 children: 15 (Total), out of which 7 are from outside the U.S, so 15 - 7 = 8 are from the U.S.
- Number of people from the U.S: 68.
- People who are either from the U.S or have 0 children:
[tex]\[ \text{Total U.S} + \text{Total with 0 children} - \text{People from the U.S and with 0 children} \][/tex]
[tex]\[ 68 + 15 - 8 = 75 \][/tex]
- Probability [tex]\(P(U.S \cup 0 \text{ children})\)[/tex] is:
[tex]\[ P(U.S \cup 0 \text{ children}) = \frac{75}{150} = 0.5 \][/tex]
#### 5. Given that the person is from the U.S, what is the probability that they have 3 or more children?
- People with 3 or more children: 19 (Total) + 75 (Total) = 94.
- From outside the U.S: 13 (with 3 children) + 41 (with 4 or more children) = 54.
- From the U.S: [tex]\(94 - 54 = 40\)[/tex].
- Given the person is from U.S, the conditional probability [tex]\(P(3 \text{ or more children} | U.S)\)[/tex] is:
[tex]\[ P(3 \text{ or more children} | U.S) = \frac{40}{68} = 0.5882 \][/tex]
#### 6. Given that the person has 1 child, what is the probability that person is from the U.S?
- Number of individuals with 1 child in total: 25.
- From within the U.S: 15.
- Given the person has 1 child, the conditional probability [tex]\(P(U.S | 1 \text{ child})\)[/tex] is:
[tex]\[ P(U.S | 1 \text{ child}) = \frac{15}{25} = 0.6 \][/tex]
### Summary
- The probability that someone is from the U.S is [tex]\(0.4533\)[/tex].
- The probability that someone has at least 2 children is [tex]\(0.7333\)[/tex].
- The probability that someone is from the U.S and has 1 child is [tex]\(0.1\)[/tex].
- The probability that someone is from the U.S or has 0 children is [tex]\(0.5\)[/tex].
- Given that the person is from the U.S, the probability that they have 3 or more children is [tex]\(0.5882\)[/tex].
- Given that the person has 1 child, the probability that person is from the U.S is [tex]\(0.6\)[/tex].
#### 1. What is the probability that someone is from the U.S?
- From the given data table:
- People from outside the U.S are [7, 10, 11, 13, 41].
- Total people are [15, 25, 16, 19, 75].
- Calculate the total number of people from outside the U.S: [tex]\(7 + 10 + 11 + 13 + 41 = 82\)[/tex].
- Calculate the total number of people: [tex]\(15 + 25 + 16 + 19 + 75 = 150\)[/tex].
- Calculate the total number of people from the U.S: [tex]\(150 - 82 = 68\)[/tex].
- Probability [tex]\(P(U.S)\)[/tex] is:
[tex]\[ P(U.S) = \frac{\text{Total people from the U.S}}{\text{Total people}} = \frac{68}{150} = 0.4533 \][/tex]
#### 2. What is the probability that someone has at least 2 children?
- Number of people with 2 or more children is: 16 (with 2 children) + 19 (with 3 children) + 75 (with 4 or more children)
- [tex]\(P(\text{At least 2 children})\)[/tex] is:
[tex]\[ P(\text{At least 2 children}) = \frac{16 + 19 + 75}{150} = \frac{110}{150} = 0.7333 \][/tex]
#### 3. What is the probability that someone is from the U.S and has one child?
- Number of people with 1 child: 25 (Total), out of which 10 are from outside the U.S, so 25 - 10 = 15 are from the U.S.
- Probability [tex]\(P(U.S \cap 1 \text{ child})\)[/tex] is:
[tex]\[ P(U.S \cap 1 \text{ child}) = \frac{15}{150} = 0.1 \][/tex]
#### 4. What is the probability that someone is from the U.S or has 0 children?
- Number of people with 0 children: 15 (Total), out of which 7 are from outside the U.S, so 15 - 7 = 8 are from the U.S.
- Number of people from the U.S: 68.
- People who are either from the U.S or have 0 children:
[tex]\[ \text{Total U.S} + \text{Total with 0 children} - \text{People from the U.S and with 0 children} \][/tex]
[tex]\[ 68 + 15 - 8 = 75 \][/tex]
- Probability [tex]\(P(U.S \cup 0 \text{ children})\)[/tex] is:
[tex]\[ P(U.S \cup 0 \text{ children}) = \frac{75}{150} = 0.5 \][/tex]
#### 5. Given that the person is from the U.S, what is the probability that they have 3 or more children?
- People with 3 or more children: 19 (Total) + 75 (Total) = 94.
- From outside the U.S: 13 (with 3 children) + 41 (with 4 or more children) = 54.
- From the U.S: [tex]\(94 - 54 = 40\)[/tex].
- Given the person is from U.S, the conditional probability [tex]\(P(3 \text{ or more children} | U.S)\)[/tex] is:
[tex]\[ P(3 \text{ or more children} | U.S) = \frac{40}{68} = 0.5882 \][/tex]
#### 6. Given that the person has 1 child, what is the probability that person is from the U.S?
- Number of individuals with 1 child in total: 25.
- From within the U.S: 15.
- Given the person has 1 child, the conditional probability [tex]\(P(U.S | 1 \text{ child})\)[/tex] is:
[tex]\[ P(U.S | 1 \text{ child}) = \frac{15}{25} = 0.6 \][/tex]
### Summary
- The probability that someone is from the U.S is [tex]\(0.4533\)[/tex].
- The probability that someone has at least 2 children is [tex]\(0.7333\)[/tex].
- The probability that someone is from the U.S and has 1 child is [tex]\(0.1\)[/tex].
- The probability that someone is from the U.S or has 0 children is [tex]\(0.5\)[/tex].
- Given that the person is from the U.S, the probability that they have 3 or more children is [tex]\(0.5882\)[/tex].
- Given that the person has 1 child, the probability that person is from the U.S is [tex]\(0.6\)[/tex].
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