Part 1: 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 employee is randomly selected from this group, what is the probability of randomly selecting an employee who eat breakfast or lunch at the office?(2 points)

Question

Part 1: 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 employee is randomly selected from this group, what is the probability of randomly selecting an employee who eat breakfast or lunch at the office?(2 points)

Responses

98
9 eighths

58
5 eighths

78
7 eighths

28
2 eighths
Question 2
Part 2: Explain how you got your answer. Show your work and use math vocabulary to explain. (2 points)

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3 answers
(click or scroll down)

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Answer 1

Bot GPT 3.5 answered
1 month ago

To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we can use the formula P(A or B) = P(A) + P(B) - P(A and B).

P(Breakfast) = 20/80 = 1/4
P(Lunch) = 50/80 = 5/8
P(Breakfast and Lunch) = 10/80 = 1/8

P(Breakfast or Lunch) = P(Breakfast) + P(Lunch) - P(Breakfast and Lunch)
= 1/4 + 5/8 - 1/8
= 2/8 + 5/8 - 1/8
= 6/8
= 3/4

Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4 or 75%.

Answer 2

bugs banny answered
1 month ago

explain pls

Answer 3

Bot GPT 3.5 answered
1 month ago

To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we calculated the individual probabilities of eating breakfast, eating lunch, and eating both breakfast and lunch.

The probability of eating breakfast is 20 employees out of 80, which is 1/4 or 0.25. The probability of eating lunch is 50 employees out of 80, which is 5/8 or 0.625. The probability of eating both breakfast and lunch is 10 employees out of 80, which is 1/8 or 0.125.

To find the probability of eating breakfast or lunch, we added the individual probabilities of eating breakfast and lunch, and then subtracted the probability of eating both breakfast and lunch (to avoid double counting). This calculation gave us 3/4 or 0.75 as the probability of randomly selecting an employee who eats breakfast or lunch at the office.

Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4 or 75%.