Question

a local bank reports that 80% of its customers maintain a checking account, 60% have a savings account, and 50% have both. a. if a customer is chosen at random, what is the probability the customer has either a checking or a savings account? (round your answer to 2 decimal places.) b. if a customer is chosen at random, what is the probability the customer does not have either a checking or a savings account? (round your answer to 2 decimal places.)

Answer 1

The Venn probabilities for this problem are given as follows:

a. 0.9 = 90%.

b. 0.1 = 10%.

What is a Venn probability?

In a Venn probability, two non-independent events are related with each other, as are their probabilities.

The "or probability" is given by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

For this problem, the events are given as follows:

• Event A: Checking account.

• Event B: Savings account.

From the text, the probabilities are given as follows:

$P(A) = 0.8, P(B) = 0.6, P(A \cap B) = 0.5$

From item a, the or probability is given as follows:

$P(A \cup B) = 0.8 + 0.6 - 0.5 = 0.9$

There is a 0.9 = 90% probability the customer has either a checking or a savings account.

For item b, none is the opposite of either, hence there is a 1 - 0.9 = 0.1 = 10% probability the customer does not have either a checking or a savings account.

More can be learned about Venn probabilities at brainly.com/question/25698611

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