Question

In a certain city, 30% of the families have a MasterCard, 20% have an American Express card, and 25% have a Visa card. Eight percent of the families have both a MasterCard and an American Express card. Twelve percent have both a Visa card and a MasterCard. Six percent have both an American Express card and a Visa card. What is the probability of selecting a family that has either a Visa card or an American Express card?

Answer 1

Answer:

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of these probabilities.

I am going to say that:

-The set A represents the probability that a family has a MasterCard.

-The set B represents the probability that a family has an American Express Card.

-The set C represents the probability that a family has a Visa card.

We have that

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

In which a represents the probability that the family only has a MasterCard, $A \cap B$ represents the probability that the family has both a MasterCard and an American Express card and $A \cap C$ represents the probability that the family has both a Master and a Visa card.

By the same logic, we also have that:

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

$C = c + (A \cap C) + (B \cap C)$

What is the probability of selecting a family that has either a Visa card or an American Express card?

This is $P = b + c$.

We start finding the probabilities from the intersection of these sets.

Six percent have both an American Express card and a Visa card. This means that:

$B \cap C = 0.06$

Twelve percent have both a Visa card and a MasterCard

$A \cap C = 0.12$

Eight percent of the families have both a MasterCard and an American Express card.

$A \cap B = 0.08$

25% have a Visa card

$C = 0.25$

$C = c + (A \cap C) + (B \cap C)$

$0.25 = c + 0.12 + 0.06$

$c = 0.07$

20% have an American Express card

$B = 0.20$

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

$0.20 = b + 0.08 + 0.06$

$b = 0.06$

What is the probability of selecting a family that has either a Visa card or an American Express card?

$P = b + c = 0.07 + 0.06 = 0.13$

There is a 13% probability of selecting a family that has either a Visa card or an American Express card