A fast-food company is interested in knowing the probability of whether a customer viewed an advertisement for their new special on the internet or on television. They found that 37% of customers saw the advertisement on the internet, 20% saw it on television, and 12% saw it on both the internet and on television. What is the probability that a randomly selected customer saw the advertisement on the internet or on television

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Answer:

45% probability that a randomly selected customer saw the advertisement on the internet or on television

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a customer saw the advertisement on the internet.

B is the probability that a customer saw the advertisement on television.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that a customer saw the advertisement on the internet but not on television and [tex]A \cap B[/tex] is the probability that the customers saw the advertisement in both the internet and on television.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

12% saw it on both the internet and on television.

This means that [tex]A \cap B = 0.12[/tex]

20% saw it on television

This means that [tex]B = 0.2[/tex]

37% of customers saw the advertisement on the internet

This means that [tex]A = 0.37[/tex]

What is the probability that a randomly selected customer saw the advertisement on the internet or on television

[tex]A \cup B = A + B - (A \cap B) = 0.37 + 0.2 - 0.12 = 0.45[/tex]

45% probability that a randomly selected customer saw the advertisement on the internet or on television