The probability that an employee at a company eats lunch at the company
cafeteria is 0.23. The probability that an employee is female is 0.52. The
probability that an employee eats lunch at the cafeteria and is female is 0.11.
What is the probability that a randomly chosen employee either eats at the
cafeteria or is female?
Let E = The randomly chosen employee eats lunch at the company cafeteria.
Let F = The randomly chosen employee is female.
We are given:
P(E) = 0.23
P(F) = 0.52
P(E and F) = 0.11
We want to find P(E or F)
We use the formula:
P(E or F) = P(E) + P(F) - P(E and F)
= 0.23 + 0.52 - 0.11
= 0.64
Edwin