12651241
contestada

The table below represents the data collected at a sandwich shop for the last six months with respect to the type of bread people ordered (sourdough or wheat) and whether or not they got cheese on their sandwich.

Are the events “sourdough” and “with cheese” independent events? Explain why or why not using probabilities. Show all work to receive full credit.

The table below represents the data collected at a sandwich shop for the last six months with respect to the type of bread people ordered sourdough or wheat and class=

Respuesta :

semsee45

Answer:

First, add totals to the table (see attachment)

Let S = Sourdough

Let C = With cheese

[tex]\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}[/tex]

[tex]\implies \sf P(S)=\dfrac{1225}{3125}=\dfrac{49}{125}[/tex]

[tex]\implies \sf P(C)=\dfrac{2000}{3125}=\dfrac{16}{25}[/tex]

[tex]\implies \sf P(S \cap C)=\dfrac{800}{3125}=\dfrac{32}{125}[/tex]

For independent events S and C,

[tex]\sf P(S \cap C)=P(S)P(C)[/tex]

[tex]\sf\:As\quad P(S) \cdot P(C)=\dfrac{49}{125} \cdot \dfrac{16}{25}=\dfrac{784}{3125}[/tex]

[tex]\sf And\quad \dfrac{32}{125} \neq \dfrac{784}{3125}[/tex]

[tex]\sf Then \quad P(S \cap C)\neq P(S)P(C)[/tex]

So the events are NOT independent

Ver imagen semsee45

Otras preguntas