select the correct answer from the drop - down menu.\nthe table shows the results from a survey of 335…

select the correct answer from the drop - down menu.\nthe table shows the results from a survey of 335 randomly selected households with pets. this survey was conducted by a new pet store that is opening nearby.\n| | have children | do not have children | total |\n|----|----|----|----|\n| 1 pet | 38 | 53 | 91 |\n| 2 pets | 85 | 41 | 126 |\n| 3 or more pets | 46 | 72 | 118 |\n| total | 169 | 166 | 335 |\nthe pet store uses the data to make decisions about inventory. complete the given statement.\na customer is more likely to have 1 pet and no children than they are to have

select the correct answer from the drop - down menu.\nthe table shows the results from a survey of 335 randomly selected households with pets. this survey was conducted by a new pet store that is opening nearby.\n| | have children | do not have children | total |\n|----|----|----|----|\n| 1 pet | 38 | 53 | 91 |\n| 2 pets | 85 | 41 | 126 |\n| 3 or more pets | 46 | 72 | 118 |\n| total | 169 | 166 | 335 |\nthe pet store uses the data to make decisions about inventory. complete the given statement.\na customer is more likely to have 1 pet and no children than they are to have

Answer

Explanation:

Step1: Calcular probabilidad de 1 mascota y sin hijos

La cantidad de clientes con 1 mascota y sin hijos es 53. La probabilidad $P(1\ mascota\ y\ sin\ hijos)$ se calcula dividiendo este número entre el total de encuestados (335). Así, $P(1\ mascota\ y\ sin\ hijos)=\frac{53}{335}$.

Step2: Calcular probabilidad de 2 mascotas y con hijos

La cantidad de clientes con 2 mascotas y con hijos es 85. La probabilidad $P(2\ mascotas\ y\ con\ hijos)=\frac{85}{335}$.

Step3: Calcular probabilidad de 3 mascotas y con hijos

La cantidad de clientes con 3 mascotas y con hijos es 46. La probabilidad $P(3\ mascotas\ y\ con\ hijos)=\frac{46}{335}$.

Step4: Calcular probabilidad de 3 mascotas y sin hijos

La cantidad de clientes con 3 mascotas y sin hijos es 72. La probabilidad $P(3\ mascotas\ y\ sin\ hijos)=\frac{72}{335}$.

Step5: Comparar probabilidades

Tenemos que $\frac{53}{335}<\frac{85}{335}$, $\frac{53}{335}<\frac{72}{335}$ y $\frac{53}{335}>\frac{46}{335}$.

Answer:

3 pets and children