In a certain town it is known that 16% of families have no children, 34% have 1 child, 23% have 2 children, 17% have 3 children, 5% have 4 children, and 5% have 5 or more children. What is the probability that a family has more than 2 children if it is known that it has at least 1 child? (Round to the nearest hundredth of a percent)

A First Course in Probability (10th Edition)
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ISBN:9780134753119
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Chapter1: Combinatorial Analysis
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**Problem Statement:**

In a certain town, it is known that:
- 16% of families have no children,
- 34% have 1 child,
- 23% have 2 children,
- 17% have 3 children,
- 5% have 4 children, and
- 5% have 5 or more children.

**Question:**
What is the probability that a family has more than 2 children if it is known that it has at least 1 child? (Round to the nearest hundredth of a percent.)

**Solution:**

First, we need to find the percentage of families that have at least 1 child:
- 34% (1 child) +
- 23% (2 children) +
- 17% (3 children) +
- 5% (4 children) +
- 5% (5 or more children) =
- Total = 84%

Next, we calculate the percentage of families that have more than 2 children:
- 17% (3 children) +
- 5% (4 children) +
- 5% (5 or more children) =
- Total = 27%

Finally, the probability that a family has more than 2 children, given it has at least 1 child, is calculated using conditional probability:
\[ P(\text{More than 2 children | At least 1 child}) = \frac{\text{% with more than 2 children}}{\text{% with at least 1 child}} = \frac{27\%}{84\%} \]

Calculate and round to the nearest hundredth of a percent:
\[ \frac{27}{84} \approx 0.3214 \] or 32.14%

Therefore, the probability is **32.14%**.
Transcribed Image Text:**Problem Statement:** In a certain town, it is known that: - 16% of families have no children, - 34% have 1 child, - 23% have 2 children, - 17% have 3 children, - 5% have 4 children, and - 5% have 5 or more children. **Question:** What is the probability that a family has more than 2 children if it is known that it has at least 1 child? (Round to the nearest hundredth of a percent.) **Solution:** First, we need to find the percentage of families that have at least 1 child: - 34% (1 child) + - 23% (2 children) + - 17% (3 children) + - 5% (4 children) + - 5% (5 or more children) = - Total = 84% Next, we calculate the percentage of families that have more than 2 children: - 17% (3 children) + - 5% (4 children) + - 5% (5 or more children) = - Total = 27% Finally, the probability that a family has more than 2 children, given it has at least 1 child, is calculated using conditional probability: \[ P(\text{More than 2 children | At least 1 child}) = \frac{\text{% with more than 2 children}}{\text{% with at least 1 child}} = \frac{27\%}{84\%} \] Calculate and round to the nearest hundredth of a percent: \[ \frac{27}{84} \approx 0.3214 \] or 32.14% Therefore, the probability is **32.14%**.
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