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Suppose in a certain state, 60% of the people have a Visa credit card, 40% have a MasterCard, and 24% have both. Let V be the event that a randomly selected individual has a Visa credit card, and M be the analogous event for a MasterCard. So P(V ) = 0.60, P(M) = 0.40, and P(V and M) = 0.24.

Respuesta :

Answer:

The question has some details missing; i.e

a. What is the probability that the shopper has neither type of card?

b. What is the probability that the shopper has both types of card?

c. What is the probability the individual has a Visa card but not a Mastercard? (Hint: You will use the answer)

a) 0.24

b) 0.24

c) 0.36

Step-by-step explanation:

The detailed steps is shown in the attachment.

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MrRoyal

Question Continuation

Given that an individual is selected at random and that he or she has at least one card, what is the probability that he or she has a Visa Card?

Answer:

0.79

Step-by-step explanation:

Given

P(V) = 0.6

P(M) = 0.4

P(V∩M)=0.24

The probability of having at least one card is:

P(V)+P(B)−P(V∩M)=0.6 + 0.4 - 0.24 = 0.76

Denote C={ at least one card }.

So, P(C) = 0.76

The probability you need (definition of conditional probability):

Denote P(V|C)= {Probability that a person with just one card has visa card}

So, P(V|C) = P(V∩C) / P(C)

If you have a Visa card, you have at least one card, so P(V∩C)=P(V) = 0.6

So, P(V|C) = P(V)/P(C)

P(V|C) = 0.6/0.76

P(V|C) = 0.78947368421

P(V|C) = 0.79 ---------- Approximated

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