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According to a survey, 22.3% of households are located in a certain region. In addition, 8.6% of households earn $75,000 or more per year and are located in the same region. Determine the probability that a randomly selected household earns $75,000 or more per year, given that the house is in the certain region. (Round your answer to one decimal place.)

User Tura
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Final answer:

The probability that a randomly selected household earns $75,000 or more per year, given that it is in a certain region, is calculated using the formula for conditional probability. The result is approximately 38.6%.

Step-by-step explanation:

The question asks to determine the probability that a randomly selected household earns $75,000 or more per year, given that the house is in a certain region.

We have two pieces of data from a survey: 22.3% of households are in that region, and 8.6% of all households earn $75,000 or more per year and are located in the same region.

To find the conditional probability, we use the definition: P(A given B) = P(A and B) / P(B).

Here, let A be the event that a household earns $75,000 or more per year, and B be the event that a household is in that region. We have P(A and B) = 8.6% and P(B) = 22.3%. So, P(A given B) = (8.6%) / (22.3%).

After calculating, we get P(A given B) = 0.3856502242 or about 38.6% when rounded to one decimal place.

Hence, the probability that a randomly selected household earns $75,000 or more per year, given that it is in the certain region, is 38.6%.

User Imbichie
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