A survey of 80 employees was conducted asking about which meals they regularly eat at the office 20 employees eat breakfast at the office 50 eat lunch and 10 eat both breakfast and lunch if an employees randomly selected from this group what is the probability of randomly selecting an employee who eats breakfast or lunch at the office? Is the answer 7/8?
the answer is 7/8
No, the answer is not 7/8.
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we can use the principle of inclusion-exclusion.
P(eat breakfast or lunch) = P(eat breakfast) + P(eat lunch) - P(eat breakfast and lunch)
P(eat breakfast or lunch) = (20/80) + (50/80) - (10/80)
P(eat breakfast or lunch) = 0.25 + 0.625 - 0.125
P(eat breakfast or lunch) = 0.75
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 0.75, not 7/8.