Question

students in a class were surveyed about the number of children in their families. the results of the survey are shown in the table. 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?
number of children in family number of surveys
one 9
two 18
three 22
four 8
five or more 3

Explanation:

Step1: Calculate total number of surveys

Add up all the number of surveys: \(9 + 18+22 + 8+3=60\)

Step2: Calculate probability of first - survey (four children)

The number of surveys with four children is \(8\). So the probability \(P_1=\frac{8}{60}=\frac{2}{15}\)

Step3: Calculate probability of second - survey (one child)

The number of surveys with one child is \(9\). So the probability \(P_2=\frac{9}{60}=\frac{3}{20}\)

Step4: Calculate the combined probability

Since the two events are independent (because the first survey is returned), use the formula \(P = P_1\times P_2\). So \(P=\frac{2}{15}\times\frac{3}{20}=\frac{6}{300}=\frac{1}{50}\)

Answer:

\(\frac{1}{50}\)