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 2/8?
No, the answer is not 2/8.
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to add the number of employees who eat breakfast to the number of employees who eat lunch, and then subtract the number of employees who eat both breakfast and lunch to avoid counting them twice.
So, the total number of employees who eat breakfast or lunch is 20 (breakfast) + 50 (lunch) - 10 (both) = 60.
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 60/80, which simplifies to 3/4.