Mathematics: A shopper is randomly selected from a market. Let A be the event the selected shopper has a Visa credit card, and B be the event the shopper has a Mastercard credit card. Suppose P(A) = 0.5, P(B) = 0.45, and P(A ∪ B) = 0.8.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 from part b.) d. Are the events

The answer will be :53/200=2,5=250%...

A shopper is randomly selected from a market. Let

24lbriscoe

23.01.2022

$1 - P(A \cup B ) = 1 - 0.8 = 0.2$

$P(A \cap B) = P(A) + P(B) - P(A \cup B ) = 0.5 + 0.45 - 0.8 = 0.15$

$P(A \cap B^c ) = P(A) - P(A \cap B) = 0.5 - 0.15 = 0.35$

They are not, the intersection is not 0.

Step-by-step explanation:

The probability that the shopper has neither type of card is the probability of the complement of the union, therefore it would be

$1 - P(A \cup B ) = 1 - 0.8 = 0.2$

That's probability of the intersection of the events. For that we use the following formula

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

Therefore

$P(A \cap B) = P(A) + P(B) - P(A \cup B ) = 0.5 + 0.45 - 0.8 = 0.15$

That's the probability of A intersection the complement of B.

For that, first of all remember that

$A = (A \cap B) \cup (A \cap B^c)$

Therefore

$P(A) = P(A \cap B) + P(A \cap B^c)$

Therefore

$P(A \cap B^c ) = P(A) - P(A \cap B) = 0.5 - 0.15 = 0.35$

They are not, the intersection is not 0.

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