To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to add the probability of selecting an employee who eats breakfast to the probability of selecting an employee who eats lunch, but we need to subtract the probability of selecting an employee who eats both breakfast and lunch to avoid double counting.
P(eats breakfast or lunch) = P(eats breakfast) + P(eats lunch) - P(eats both)
P(eats breakfast) = 20/80 = 0.25
P(eats lunch) = 50/80 = 0.625
P(eats both) = 15/80 = 0.1875
P(eats breakfast or lunch) = 0.25 + 0.625 - 0.1875 = 0.6875
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 0.6875 or 68.75%.
A survey of 80 employees was conducted asking about which meals they regularly eat at the office. Twenty employees eat breakfast at the office, 50 eat lunch, and 15 eat both breakfast and lunch. If an employee is randomly selected from this group, what is the probability of randomly selecting an employee who eat breakfast or lunch at the office? Explain how you got the answer.
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