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AlexFokin [52]
4 years ago
10

The table gives the relative frequencies of recipes that contains sugar and salt, contain at least one of those ingredients, or

contains neither of those ingredients
what is the probability that a random selected recipe does not contain sugar, given that it contains salt

write the probability as a percent. round to the nearest tenth if needed

https://imgur.com/a/gqNBhip
Mathematics
1 answer:
tigry1 [53]4 years ago
5 0

<u>Answer-</u>

<em>The probability that a randomly selected recipe does not contain sugar, given that it contains salt is 22.4%</em>

<u>Solution-</u>

The given table in the link shows the relative frequencies of recipes that contains sugar and salt, or contains at least one of those ingredients, or contains neither of those ingredients.

We have to find the conditional probability that the recipe doesn't contain sugar, given that it contains salt.

We know that, the conditional probability of occurrence of A given that B occurs is,

P(A|B)=\frac{P(A\ and\ B)}{P(B)} =\frac{P(A\bigcap B)}{P(B)}

P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})}{P(\text{Contains salt})}


P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})=\frac{0.15}{1} =0.15\\P(\text{Contains salt})=\frac{0.67}{1}=0.67

Putting these values,

P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{0.15}{0.67} =0.224=22.4\%

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Vamos a utilizar el método de sustitución. Vamos a despejar la segunda ecuación por alguna de las variables. Elegiré la y porque esta sola.

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Listo.

------------------------------------------------------------------------

Procedemos a reemplazar el valor de y en la primera ecuación.

\frac{5x+2y}{3}=1

\frac{5x+2(2-2x)}{3}=1

\frac{5x+4-4x}{3} =1

Procedemos a operar términos semejantes.

\frac{x+4}{3} =1

Nos deshacemos del 3 multiplicando.

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Restamos 4

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-----------------------------------------------------------------------------

Una vez encontrado el valor de x, procedemos a reemplazar en cualquier ecuación para encontrar el valor de y. Usaré la segunda ecuación.

\frac{2x+y}{2}=1

\frac{2(-1)+y}{2}=1

Procedemos a resolver;

\frac{-2+y}{2} =1

Multiplicamos por 2.

-2+y=1*2

-2+y=2

Sumamos 2

y=2+2\\y=4

-------------------------------------------------------------------------------

Para comprobar que los valores son correctos, reemplazamos ambos valores en cualquier ecuación y el resultado debe ser igual.

Usaré la primera.

\frac{5x+2y}{3} =1

\frac{5(-1)+2(4)}{3} =1

Resolvemos;

\frac{-5+8}{3} =1

\frac{3}{3}=1

1=1

6 0
4 years ago
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