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Sagot :
Let's address each question one by one using the provided data.
1. What is the probability that someone is from the U.S?
There are 75 people in total, and 34 of them are from the U.S.
[tex]\[ P(\text{from the U.S}) = \frac{\text{Number of people from the U.S}}{\text{Total number of people}} = \frac{34}{75} \approx 0.4533 \][/tex]
2. What is the probability that someone has at least 2 children?
The number of people with at least 2 children is the sum of those with 2 children and those with 3 or more children.
[tex]\[ \text{Number of people with at least 2 children} = 16 + 19 = 35 \][/tex]
[tex]\[ P(\text{at least 2 children}) = \frac{\text{Number of people with at least 2 children}}{\text{Total number of people}} = \frac{35}{75} \approx 0.4667 \][/tex]
3. What is the probability that someone is from the U.S and has one child?
There are 15 people from the U.S. who have one child.
[tex]\[ P(\text{from the U.S and one child}) = \frac{\text{Number of people from the U.S with one child}}{\text{Total number of people}} = \frac{15}{75} = 0.2 \][/tex]
4. What is the probability that someone is from the U.S or has 0 children?
This involves the union of two events: being from the U.S and having 0 children. We use the formula for the union of two probabilities:
[tex]\[ P(\text{U.S}) = 0.4533 \][/tex]
[tex]\[ P(\text{0 children}) = \frac{15}{75} = 0.2 \][/tex]
There are 8 people from the U.S who have 0 children.
[tex]\[ P(\text{U.S and 0 children}) = \frac{8}{75} \][/tex]
Using the formula for the union of two events:
[tex]\[ P(\text{U.S or 0 children}) = P(\text{U.S}) + P(\text{0 children}) - P(\text{U.S and 0 children}) \][/tex]
[tex]\[ = 0.4533 + 0.2 - \frac{8}{75} \][/tex]
[tex]\[ = 0.4533 + 0.2 - 0.1067 \approx 0.44 \][/tex]
5. Given that the person is from the U.S, what is the probability that they have 3 or more children?
We need the conditional probability [tex]\(P(3 \text{ or more children} | \text{U.S})\)[/tex].
[tex]\[ P(3 \text{ or more children} | \text{U.S}) = \frac{\text{Number of people from the U.S with 3 or more children}}{\text{Total number of people from the U.S}} = \frac{6}{34} \approx 0.1765 \][/tex]
6. Given that the person has 1 child, what is the probability that person is from the U.S?
We need the conditional probability [tex]\(P(\text{U.S} | 1 \text{ child})\)[/tex].
[tex]\[ P(\text{U.S} | 1 \text{ child}) = \frac{\text{Number of people from the U.S with 1 child}}{\text{Total number of people with 1 child}} = \frac{15}{25} = 0.6 \][/tex]
Thus, the answers are:
1. 0.4533
2. 0.4667
3. 0.2
4. 0.44
5. 0.1765
6. 0.6
1. What is the probability that someone is from the U.S?
There are 75 people in total, and 34 of them are from the U.S.
[tex]\[ P(\text{from the U.S}) = \frac{\text{Number of people from the U.S}}{\text{Total number of people}} = \frac{34}{75} \approx 0.4533 \][/tex]
2. What is the probability that someone has at least 2 children?
The number of people with at least 2 children is the sum of those with 2 children and those with 3 or more children.
[tex]\[ \text{Number of people with at least 2 children} = 16 + 19 = 35 \][/tex]
[tex]\[ P(\text{at least 2 children}) = \frac{\text{Number of people with at least 2 children}}{\text{Total number of people}} = \frac{35}{75} \approx 0.4667 \][/tex]
3. What is the probability that someone is from the U.S and has one child?
There are 15 people from the U.S. who have one child.
[tex]\[ P(\text{from the U.S and one child}) = \frac{\text{Number of people from the U.S with one child}}{\text{Total number of people}} = \frac{15}{75} = 0.2 \][/tex]
4. What is the probability that someone is from the U.S or has 0 children?
This involves the union of two events: being from the U.S and having 0 children. We use the formula for the union of two probabilities:
[tex]\[ P(\text{U.S}) = 0.4533 \][/tex]
[tex]\[ P(\text{0 children}) = \frac{15}{75} = 0.2 \][/tex]
There are 8 people from the U.S who have 0 children.
[tex]\[ P(\text{U.S and 0 children}) = \frac{8}{75} \][/tex]
Using the formula for the union of two events:
[tex]\[ P(\text{U.S or 0 children}) = P(\text{U.S}) + P(\text{0 children}) - P(\text{U.S and 0 children}) \][/tex]
[tex]\[ = 0.4533 + 0.2 - \frac{8}{75} \][/tex]
[tex]\[ = 0.4533 + 0.2 - 0.1067 \approx 0.44 \][/tex]
5. Given that the person is from the U.S, what is the probability that they have 3 or more children?
We need the conditional probability [tex]\(P(3 \text{ or more children} | \text{U.S})\)[/tex].
[tex]\[ P(3 \text{ or more children} | \text{U.S}) = \frac{\text{Number of people from the U.S with 3 or more children}}{\text{Total number of people from the U.S}} = \frac{6}{34} \approx 0.1765 \][/tex]
6. Given that the person has 1 child, what is the probability that person is from the U.S?
We need the conditional probability [tex]\(P(\text{U.S} | 1 \text{ child})\)[/tex].
[tex]\[ P(\text{U.S} | 1 \text{ child}) = \frac{\text{Number of people from the U.S with 1 child}}{\text{Total number of people with 1 child}} = \frac{15}{25} = 0.6 \][/tex]
Thus, the answers are:
1. 0.4533
2. 0.4667
3. 0.2
4. 0.44
5. 0.1765
6. 0.6
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