A basket of fruits contains 20 apples, 4 watermelon slices, and 12 oranges. If

Question

a piece of fruit is to be selected randomly from the box, what is the probability
of selecting an apple or an orange? Explain the process you used.

Answer 1

The process I used is to add the number of apples and oranges together, since we want to find the probability of selecting either one of these fruits.

Total number of apples and oranges = 20 + 12 = 32

The probability of selecting an apple or an orange is therefore:

P(apple or orange) = (number of apples and oranges) / (total number of fruits)

P(apple or orange) = 32 / (20 + 4 + 12)

P(apple or orange) = 32 / 36

P(apple or orange) = 0.89 or 89%

Therefore, the probability of selecting an apple or an orange is 0.89 or 89%.

Answer 2

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.

Answer 3

To calculate the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to use the formula:

P(A or B) = P(A) + P(B) - P(A and B)

Where A = employees who eat breakfast at the office, B = employees who eat lunch at the office.

We are given that:

- P(A) = 20/80 = 1/4
- P(B) = 50/80 = 5/8
- P(A and B) = 15/80

Substituting these values into the formula:

P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = 1/4 + 5/8 - 15/80
P(A or B) = 20/80 + 50/80 - 15/80
P(A or B) = 55/80

Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 55/80 or 0.6875.