At a certain company, 40% of the employees are certified to operate machine A, 5

Question

At a certain company, 40% of the employees are certified to operate machine A, 50% are certified to operate machine B, 40% are certified to operate machine C, 15% are certified to operate machines A and B, 10% are certified to operate machines A and C, 5% are certified to operate all three machines, and 15% are certified to operate machine C but neither machine A nor machine B. Let A denote the event that a randomly chosen employee is certified to operate machine A, let B denote the event that a randomly chosen employee is certified to operate machine B, and let C denote the event that a randomly chosen employee is certified to operate machine C. For each of the following, include an expression for the event in terms of set operations on A, B and C, and justify your answer.

(a) Find the probability that a randomly chosen employee is certified to operate at exactly one of the three machines.

(b) Find the probability that a randomly chosen employee is certified to operate machine A and machine C but not machines B.

(c) Find the probability that a randomly chosen employee is certified to operate two of the three machines but not all three.

(d) Find the probability that a randomly chosen employee is certified to operate at least one of the three machines

Explanation / Answer

P(A) = 0.4

P(B) = 0.5

P(C) = 0.4

P(A and B) = 0.15

P(A and C) = 0.1

P(A and B and C) = 0.05

P(only C) = 0.15

P(only C) = P(C) - P(A and C) - P(B and C) + P(A and B and C)

Or, 0.15 = 0.4 - 0.1 - P(B and C) + 0.05

Or, P(B and C) = 0.2

A) P(only A) = P(A) - P(A and B) - P(A and C) + P(all three)

= 0.4 - 0.15 - 0.1 + 0.05

= 0.2

P(only B) = 0.5 - 0.15 - 0.2 + 0.05 = 0.2

P(Exactly one machine) = P(only A) + P(only B) + P(only C)

= 0.2 + 0.2 + 0.15

= 0.55

B) P(A and C and not B) = P(A and C) - P(all three)

= 0.1 - 0.05

= 0.05

C) P(A and B and not C) = P(A and B) - P(all three) = 0.15 - 0.05 = 0.1

P(B and C and not A) = 0.2 - 0.05 = 0.15

D) P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P(all three)

= 0.4 + 0.5 + 0.4 - 0.15 - 0.1 - 0.2 + 0.05

= 0.9

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