Answered

Students in a class were surveyed about the number of children in their families. The results of the survey are shown in the table.




Number of Children in Family		Number of Surveys

one		9

two		18

three		22

four		8

five or more		3






Two surveys are chosen at random from the group of surveys. After the first survey is chosen, it is returned to the stack and can be chosen a second time. What is the probability that the first survey chosen indicates four children in the family and the second survey indicates one child in the family?


A$\frac{1}{50}$
B$\frac{2}{15}$
C$\frac{3}{20}$
D$\frac{17}{60}$

Verified Answer :

  • Total number of surveys is \(N = 60\).
  • Probability of selecting a 'four children' survey is \(P(\text{four}) = \frac{8}{60} = \frac{2}{15}\).
  • Probability of selecting a 'one child' survey is \(P(\text{one}) = \frac{9}{60} = \frac{3}{20}\).
  • Since the events are independent, the joint probability is \(\frac{2}{15} \times \frac{3}{20} = \boxed{\frac{1}{50}}\).

Explanation

  1. Calculate the total number of surveys

To determine the probability of selecting a specific sequence of surveys, we first need to identify the total number of surveys collected. We are given the number of surveys for each category of the number of children in a family:

  • One child: 9 surveys
  • Two children: 18 surveys
  • Three children: 22 surveys
  • Four children: 8 surveys
  • Five or more children: 3 surveys

Adding these values gives the total number of surveys (\(N\)):
\[N = 9 + 18 + 22 + 8 + 3 = 60\]

  1. Determine the probabilities of individual events

Next, we identify the number of surveys that fit our specific criteria.

  • Number of surveys indicating 'four children': \(n_4 = 8\)
  • Number of surveys indicating 'one child': \(n_1 = 9\)

Since the surveys are chosen with replacement, the probability of selecting a specific category remains constant for each draw. The probability (\(P\)) of selecting a survey indicating 'four children' in the first draw is:
\[P(\text{four}) = \frac{n_4}{N} = \frac{8}{60} = \frac{2}{15}\]
The probability of selecting a survey indicating 'one child' in the second draw is:
\[P(\text{one}) = \frac{n_1}{N} = \frac{9}{60} = \frac{3}{20}\]

  1. Calculate the combined probability

Because the first survey is returned to the stack before the second survey is chosen, the two events are independent. To find the probability of both events occurring in the specified order (first 'four', then 'one'), we use the multiplication rule for independent events:
\[P(\text{four then one}) = P(\text{four}) \times P(\text{one})\]
\[P(\text{four then one}) = \frac{2}{15} \times \frac{3}{20} = \frac{6}{300}\]
Simplifying this fraction:
\[\frac{6}{300} = \frac{1}{50}\]
Therefore, the probability is \(\frac{1}{50}\).

Examples

This type of probability calculation is essential in data analysis and quality control, such as when a manager needs to estimate the likelihood of specific customer segments being sampled consecutively in a survey. By understanding sampling with replacement, one can predict the frequency of specific outcomes in repeated random observations, which helps in making informed decisions about resource allocation or marketing strategies. For example, if you know the composition of a group, you can calculate the odds of drawing specific items in a set sequence, helping you set realistic expectations for your data collection results.

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