To solve this problem, we need to find the probability that two surveys chosen at random (with replacement) will indicate that the first survey indicates four children in the family and the second survey indicates one child in the family.
Let's break down the solution step-by-step:
1. Calculate the total number of surveys:
The total number of surveys [tex]\( N \)[/tex] is the sum of all the surveys conducted:
[tex]\[
N = 9 + 18 + 22 + 8 + 3 = 60
\][/tex]
2. Find the probability of selecting a survey that indicates four children:
The number of surveys indicating four children is [tex]\( 8 \)[/tex]. Thus, the probability [tex]\( P(\text{four children}) \)[/tex] is:
[tex]\[
P(\text{four children}) = \frac{8}{60} = \frac{2}{15} \approx 0.1333
\][/tex]
3. Find the probability of selecting a survey that indicates one child:
The number of surveys indicating one child is [tex]\( 9 \)[/tex]. Thus, the probability [tex]\( P(\text{one child}) \)[/tex] is:
[tex]\[
P(\text{one child}) = \frac{9}{60} = \frac{3}{20} = 0.15
\][/tex]
4. Since the survey is returned to the stack, the events are independent:
To find the combined probability of these two independent events occurring in sequence, we multiply their individual probabilities:
[tex]\[
P(\text{four children and then one child}) = P(\text{four children}) \times P(\text{one child}) = \left( \frac{2}{15} \right) \times \left( \frac{3}{20} \right)
\][/tex]
5. Calculate the combined probability:
[tex]\[
P(\text{four children and then one child}) = \frac{2}{15} \times \frac{3}{20} = \frac{2 \times 3}{15 \times 20} = \frac{6}{300} = \frac{1}{50}
\][/tex]
Therefore, the probability that the first survey chosen indicates four children in the family and the second survey indicates one child in the family is:
[tex]\[
\boxed{\frac{1}{50}}
\][/tex]
