Question

Probability

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)??

Answer 1

Fraction please

Answer 2

Over 8

Answer 3

To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to use the principle of inclusion-exclusion.

Let A be the event of selecting an employee who eats breakfast at the office, and B be the event of selecting an employee who eats lunch at the office.

P(A) = 20/80 = 0.25
P(B) = 50/80 = 0.625
P(A and B) = 10/80 = 0.125 (employees who eat both)

Now, the probability of selecting an employee who eats breakfast or lunch can be calculated using the formula:

P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 0.25 + 0.625 - 0.125
P(A or B) = 0.75

Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 0.75 or 75%.

Answer 4

The probability, as a fraction, of randomly selecting an employee who eats breakfast or lunch at the office is 3/4.

Answer 5

The probability, as a fraction, of randomly selecting an employee who eats breakfast or lunch at the office is 6/8, which can be simplified to 3/4.