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

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