Question

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

Answer 1

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:

$A = a + (A \cap B)$

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

By the same logic, we have that:

$B = b + (A \cap B)$

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

This means that $A \cap B = 0.12$

20% saw it on television

This means that $B = 0.2$

37% of customers saw the advertisement on the internet

This means that $A = 0.37$

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

$A \cup B = A + B - (A \cap B) = 0.37 + 0.2 - 0.12 = 0.45$

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