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Suppose that 65% of Americans over 18 drink coffee in the morning, 25% of Americans over the age of 18 have cereal for breakfast, and 10% do both. What is the probability that a randomly selected american over the age of 18 drinks coffee in the morning or has cereal for breakfast? That is, find P(C or B).

User Rgchris
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3 votes

Answer:

c

Explanation:

User Matthew Moss
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4 votes

Explanation:

To find the probability that a randomly selected American over the age of 18 drinks coffee in the morning or has cereal for breakfast, we can use the formula:

P(C or B) = P(C) + P(B) - P(C and B)

where:

P(C) = the probability of drinking coffee in the morning

P(B) = the probability of having cereal for breakfast

P(C and B) = the probability of doing both

From the problem, we know that:

P(C) = 0.65

P(B) = 0.25

P(C and B) = 0.10

Plugging these values into the formula, we get:

P(C or B) = 0.65 + 0.25 - 0.10

P(C or B) = 0.80

Therefore, the probability that a randomly selected American over the age of 18 drinks coffee in the morning or has cereal for breakfast is 0.80, or 80%.

User GManNickG
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