Question

algebra i b-cr
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?
○ $\frac{1}{50}$
○ $\frac{2}{15}$
○ $\frac{3}{20}$
○ $\frac{17}{60}$
mark this and return save and exit next submit

Explanation:

Step1: Calculate total surveys

$9 + 18 + 22 + 8 + 3 = 60$

Step2: Find P(first=4 children)

$\frac{\text{Number of 4-child surveys}}{\text{Total surveys}} = \frac{8}{60} = \frac{2}{15}$

Step3: Find P(second=1 child)

$\frac{\text{Number of 1-child surveys}}{\text{Total surveys}} = \frac{9}{60} = \frac{3}{20}$

Step4: Multiply the two probabilities

$\frac{2}{15} \times \frac{3}{20} = \frac{6}{300} = \frac{1}{50}$

Answer:

$\frac{1}{50}$ (Option A)