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Use the table below to answer the following questions:

\begin{tabular}{|r|r|r|r|r|r|}
\hline
\# of Children & 0 & 1 & 2 & 3 or more & Total \\
\hline
From the U.S & 8 & 15 & 5 & 6 & 34 \\
\hline
From outside the U.S & 7 & 10 & 11 & 13 & 41 \\
\hline
Total & 15 & 25 & 16 & 19 & 75 \\
\hline
\end{tabular}

1. What is the probability that someone is from the U.S? [tex]$\square$[/tex]

2. What is the probability that someone has at least 2 children? [tex]$\square$[/tex]

3. What is the probability that someone is from the U.S and has one child? [tex]$\square$[/tex]


Sagot :

Certainly! Let's go through each of these questions step-by-step using the data provided in the table to determine the probabilities.

### 1. Probability that someone is from the U.S.

The total number of people surveyed is [tex]\(75\)[/tex]. Out of these, [tex]\(34\)[/tex] are from the U.S.

The probability [tex]\(P(\text{from the U.S})\)[/tex] can be calculated as:
[tex]\[ P(\text{from the U.S}) = \frac{\text{Number of people from the U.S}}{\text{Total number of people}} = \frac{34}{75} \][/tex]

Simplifying this fraction gives:
[tex]\[ P(\text{from the U.S}) \approx 0.4533 \][/tex]

### 2. Probability that someone has at least 2 children.

Let's first calculate the total number of people who have at least 2 children. This includes those with [tex]\(2\)[/tex] children and those with [tex]\(3\)[/tex] or more children.

From the table:
- Number of people with [tex]\(2\)[/tex] children: [tex]\(16\)[/tex]
- Number of people with [tex]\(3\)[/tex] or more children: [tex]\(19\)[/tex]

Summing these gives the total number of people with at least 2 children:
[tex]\[ 16 + 19 = 35 \][/tex]

The probability [tex]\(P(\text{at least 2 children})\)[/tex] can be calculated as:
[tex]\[ P(\text{at least 2 children}) = \frac{\text{Number of people with at least 2 children}}{\text{Total number of people}} = \frac{35}{75} \][/tex]

Simplifying this fraction gives:
[tex]\[ P(\text{at least 2 children}) \approx 0.4667 \][/tex]

### 3. Probability that someone is from the U.S. and has one child.

From the table, the number of people from the U.S. who have one child is [tex]\(15\)[/tex].

The probability [tex]\(P(\text{from the U.S. and has one child})\)[/tex] can be calculated as:
[tex]\[ P(\text{from the U.S. and has one child}) = \frac{\text{Number of U.S. people with one child}}{\text{Total number of people}} = \frac{15}{75} \][/tex]

Simplifying this fraction gives:
[tex]\[ P(\text{from the U.S. and has one child}) = 0.2 \][/tex]

### Summary of Probabilities:
1. The probability that someone is from the U.S. is approximately [tex]\(0.4533\)[/tex].
2. The probability that someone has at least 2 children is approximately [tex]\(0.4667\)[/tex].
3. The probability that someone is from the U.S. and has one child is [tex]\(0.2\)[/tex].