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.

number of children in family	number of surveys
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?

○ $\frac{1}{80}$
○ $\frac{2}{18}$
○ $\frac{2}{90}$
○ $\frac{1}{60}$

Explanation:

Step1: Calculate total number of surveys

First, find the sum of all the number - of - surveys values: $9 + 18+22 + 8+3=60$.

Step2: Calculate probability of first event

The probability that the first survey indicates four children in the family is $P_1=\frac{8}{60}$ since there are 8 surveys with four children out of 60 total surveys.

Step3: Calculate probability of second event

The probability that the second survey indicates one child in the family is $P_2=\frac{9}{60}$ since there are 9 surveys with one child out of 60 total surveys.

Step4: Calculate joint - probability

Since the first survey is replaced before the second draw, the two events are independent. For independent events $A$ and $B$, $P(A\cap B)=P(A)\times P(B)$. So the probability that the first survey indicates four children and the second survey indicates one child is $P = P_1\times P_2=\frac{8}{60}\times\frac{9}{60}=\frac{72}{3600}=\frac{1}{50}$.

Answer:

$\frac{1}{50}$