Question

at a certain non - profit organization, 62% of employees are college graduates and 40% of employees have more than ten years of experience. if 37% of the organization’s employees both have more than ten years of experience and are college graduates, what is the probability that a randomly selected employee will be either a college graduate or have more than ten years of experience (or both)? write your answer as a decimal (not as a percentage).

Explanation:

Step1: Recall the principle of inclusion - exclusion for probabilities.

The formula for \( P(A \cup B) \) is \( P(A) + P(B) - P(A \cap B) \), where \( A \) is the event that an employee is a college graduate and \( B \) is the event that an employee has more than ten years of experience.
Let \( P(A)=0.62 \) (probability of being a college graduate), \( P(B) = 0.40\) (probability of having more than ten years of experience), and \( P(A\cap B)=0.37 \) (probability of being both a college graduate and having more than ten years of experience).

Step2: Apply the inclusion - exclusion formula.

Substitute the values into the formula: \( P(A\cup B)=P(A)+P(B)-P(A\cap B) \)
\( P(A\cup B)=0.62 + 0.40- 0.37 \)
First, add \( 0.62\) and \( 0.40\): \( 0.62 + 0.40=1.02 \)
Then subtract \( 0.37\) from the result: \( 1.02-0.37 = 0.65 \)

Answer:

\( 0.65 \)