To solve this problem, we need to understand the probabilities of selecting specific surveys when surveys are drawn with replacement.
First, let's determine the total number of surveys conducted. Adding up all the surveys:
[tex]\[ 9 (one) + 18 (two) + 22 (three) + 8 (four) + 3 (five or more) = 60 \][/tex]
Thus, the total number of surveys conducted is 60.
Next, let's find the probability of choosing a survey indicating four children and then another survey indicating one child, given the surveys are replaced.
### Step 1: Probability of choosing a survey indicating four children
There are 8 surveys that indicate four children out of a total of 60 surveys. Thus, the probability [tex]\( P(\text{four children}) \)[/tex] is:
[tex]\[ P(\text{four children}) = \frac{8}{60} = \frac{2}{15} \][/tex]
### Step 2: Probability of choosing a survey indicating one child
There are 9 surveys that indicate one child out of a total of 60 surveys. Thus, the probability [tex]\( P(\text{one child}) \)[/tex] is:
[tex]\[ P(\text{one child}) = \frac{9}{60} = \frac{3}{20} \][/tex]
### Step 3: Combined probability with replacement
Since the surveys are replaced, these two events are independent. The combined probability [tex]\( P(\text{four children and one child}) \)[/tex] is the product of the individual probabilities:
[tex]\[ P(\text{four children and one child}) = P(\text{four children}) \times P(\text{one child}) \][/tex]
[tex]\[ P(\text{four children and one child}) = \left( \frac{2}{15} \right) \times \left( \frac{3}{20} \right) = \frac{2 \times 3}{15 \times 20} = \frac{6}{300} = \frac{1}{50} \][/tex]
Hence, the probability that the first survey chosen indicates four children and the second survey indicates one child is:
[tex]\[ \boxed{\frac{1}{50}} \][/tex]
