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15)    Sixty-four percent (64%) of the employees at a business have a Business degree.  Seventy-five percent (75%) of the employees at the same business have an MBA.  If 53% have both a Business degree and an MBA, what is the probability that a randomly chosen individual has either Business degree OR an MBA?

Question

15)    Sixty-four percent (64%) of the employees at a business have a Business degree.  Seventy-five percent (75%) of the employees at the same business have an MBA.  If 53% have both a Business degree and an MBA, what is the probability that a randomly chosen individual has either Business degree OR an MBA?

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Solution

To find the probability that a randomly chosen individual has either a Business degree or an MBA, we can use the principle of inclusion and exclusion. The formula for the probability of either event A or event B occurring is:

P(A∪B)=P(A)+P(B)−P(A∩B) P(A \cup B) = P(A) + P(B) - P(A \cap B)

Where:

  • P(A) P(A) is the probability of having a Business degree.
  • P(B) P(B) is the probability of having an MBA.
  • P(A∩B) P(A \cap B) is the probability of having both a Business degree and an MBA.

Given:

  • P(A)=0.64 P(A) = 0.64
  • P(B)=0.75 P(B) = 0.75
  • P(A∩B)=0.53 P(A \cap B) = 0.53

Now, substitute these values into the formula:

P(A∪B)=0.64+0.75−0.53 P(A \cup B) = 0.64 + 0.75 - 0.53

Calculate the result:

P(A∪B)=1.39−0.53 P(A \cup B) = 1.39 - 0.53 P(A∪B)=0.86 P(A \cup B) = 0.86

Therefore, the probability that a randomly chosen individual has either a Business degree or an MBA is 0.86, or 86%.

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